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c Notes: F.P. Greenleaf 2003 - 2010
v43-f10products.tex, version 12/1/10
Algebra I: Chapter 6. The structure of groups. 6.1 Direct products of groups.
We begin with a basic product construction. 6.1.1 Definition (External Direct Product). Given groups A1 , . . . , An we define their external direct product to be the Cartesian product set G = A1 . . .An equipped with component-by-component multiplication of n-tuples. If a = (a1 , . . . , an ), b = (b1 , . . . , bn ) in the Cartesian product set G, their product is (1) a b = (a1 , . . . , an ) (b1 , . . . , bn ) = (a1 b1 , . . . , an bn ) for all ai , bi Ai
The identity element is e = (e1 , . . . , en ) where ei is the identity element in Ai ; the inverse of an element is a-1 = (a-1 , . . . , a-1 ). n 1 There is a natural isomorphism between Ai and the subgroup Ai = (e1 ) . . . Ai . . . (en ) , the n-tuples whose entries are trivial except for ai . From (1) it is clear that (a) Each Ai is a subgroup in G. (b) The bijective map Ji (ai ) = (e1 , . . . , ai , . . . , en ) defines an isomorphism from Ai to Ai . (c) The Ai commute with each other in the sense that xy = yx if x Ai , y Aj and i = j. (d) Each Ai is a normal subgroup in G. (e) The product set A1 . . . An = {x1 . . . xn : xi Ai , 1 i n} is all of G. Note carefully what (c) does not say: the subgroup Ai need not commute with itself (the case when i = j) unless the group Ai happens to be abelian. The subsets H i = A1 . . . Ai-1 (ei ) Ai+1 . . . An G are also normal subgroups, and in a group-theoretic sense the H i are complementary to the Ai . We have the following properties. (i) H i Ai = (e) (2) (ii) G = A1 A2 . . . An (product of subsets in G) (iii) Each complement H i is a normal subgroup in G (iv) Ai G/H i via the bijection fi : ai Ji (ai )H i =
6.1.2 Exercise. Verify the claims (a) (e) regarding the subgroups Ai in a direct product G = A1 . . . An . 6.1.3 Exercise. Verify the relations (2) between the subgroups Ai = Ai and their complementary subgroups H i . 6.1.4 Exercise. Verify that the map fi : Ai G/H i defined in (iii) above is actually a 1
bijection, and that it is a homomorphism from Ai to the quotient group G/H i , so that G/H i Ai . = The order of entries in an n-tuple makes a difference; therefore the Cartesian product sets A1 A2 and A2 A1 are not the same thing (unless A1 = A2 ). For instance, are the direct product groups Z3 Z5 and Z5 Z3 the same? What do elements in these groups look like? However, in dealing with groups we only care whether they are isomorphic. It happens that Z3 Z5 Z5 Z3 even though these groups are not "identical." = 6.1.5 Exercise. Let A1 , A2 , . . . be groups. Prove that the following product groups are isomorphic. (a) A1 A2 A2 A1 = (b) A1 (A2 A3 ) (A1 A2 ) A3 A1 A2 A3 (as defined in (1)) = = (c) If one of the groups is trivial we get A1 (e2 ) A1 = (e1 ) A2 A2 =
In essence, the operation of forming the direct product of two groups is commutative and associative, and the trivial group E = (e) acts as an "identity element." The significance of (b) is that, up to isomorphism, we get the same group if we multiply groups together all at once, as in (1), or multiply them successively two at a time. 6.1.6 Exercise. In the product group G = A A the diagonal is the subset = {(a, a) : a A} (a) Show that is a subgroup of G (b) Show that = G. (c) Find a complete set of coset representatives for the coset space G/. In general is not a normal subgroup of G G. Can you see why? In (c) you are looking for a set X G that meets each coset g in a unique point. Such a set is referred to as a transversal for the space of cosets G/. 6.1.7 Example. Euclidean space Rn equipped with vector addition (+) as a binary operation is an abelian group. Its elements are defined as n-tuples x = (x1 , . . . , xn ) of real numbers, elements of the Cartesian product set R . . . R. Comparison of the sum x + y = (x1 + y1 , . . . , xn + yn ) with (1) shows that G = (Rn , +) is precisely the direct product group (R . . . R, +) made up of n copies of the real line (R, +). 6.1.8 Exercise. The set of integral vectors in Rn Zn = {x Rn : x = (x1 , . . . , xn ) with xi Z for all i} is a group under vector addition (+). Explain why (Zn , +) Z . . . Z. If {ui } is an = R-basis for Rn and = {a1 u1 + . . . + an un : ai Z} = Zu1 + . . . + Zun prove that (, +) is an abelian group isomorphic to (Zn , +). What map : Zn effects the isomorphism? There is an important "internal" version of the direct product construction. Given a group G and subgroups Ai G we would like to know when G is isomorphic to the 2
direct product A1 . . .An . From (2) we can read out some obvious necessary conditions, (i) Each Ai must be a normal subgroup in G. (3A) (ii) The Ai must generate G in the sense that G is equal to the product set A1 An = {x1 xn : xi Ai } Furthermore it is evident from the definition of the direct product G = A1 . . . An that each group element g has a unique decomposition as a product g = x1 xn with xi Ai . That imposes a third condition (3B) (iii) Each g G has a unique factorization g = x1 xn such that xi Ai . Note that (ii) insures the existence of such a factorization while (iii) insures its uniqueness. Conditions (i) (iii) are also sufficient, but before we prove that we must work out a few simple consequences of these hypotheses. 6.1.9 Exercise. Let G be a group and A, B two subgroups. Then the product set AB = {ab : a A, b B} is a subgroup if either A or B is normal, and the product is a normal subgroup if both A and B are normal. Note: Recall the discussion of Section 3.3, especially Exercise 3.3.14. 6.1.10 Lemma. Let G be a group and A1 , . . . , An subgroups that satisfy conditions (i) (iii) set forth in (3). Then the "complementary sets" Hi =
j=i
Aj = A1 . . . Ai-1 ei Ai+1 . . . An
are normal subgroups that have the following "disjointness" property. (4) Hi Ai = (e) for each 1 i n
Proof: By 6.1.9 the product AB = {ab : a A, b B} of two normal subgroups is again a normal subgroup, hence Hi is normal, being the product of several normal subgroups. As for disjointness, any g Hi Ai can be decomposed two ways g = e . . . e ai e . . . e = a1 . . . ai-1 e ai+1 . . . an These must agree. Unique factorization forces aj = e for all j, so g = e and the intersection is trivial. Note: Except in the special case when there are just two factors, statement (4) is much stronger than "pairwise disjointness" Ai Aj = (e) of the subgroups Ai when i = j. An analogous situation arises in linear algebra when we ask whether a family of vector subspaces V1 , . . . , Vr V is "linearly independent." This does not follow from the pairwise disjointness condition Vi Vj = (0), unless there are just two subspaces; instead, we must require that each Vi have trivial intersection with the linear span Hi = j=i Vj of all the other subspaces. As a simple example, consider three lines V1 , V2 , V3 (one-dimensional subspaces) in R3 . These need not be linearly independent if they are merely pairwise disjoint in the sense that Vi Vj = (0) for i = j; they might, for instance, all lie in the xy-plane. But they are linearly independent if each Vi is essentially disjoint from the 2-dimensional subspace Hi spanned by the other two lines. 6.1.11 Lemma. Let G be a group and A1 , . . . , An subgroups that satisfy conditions (i) and (ii) of (3A). Then the following statements are equivalent (iii) unique factorization: Each g G can be written uniquely as a product g = x1 xn with xi Ai . (iii) disjointness condition: The complementary subgroups Hi = have the disjointness property Ai Hi = (e) for each 1 i n. 3
j=i
Aj
Proof: We've already proved (iii) (iii) in 6.1.10. To prove (iii) (iii), suppose some g has distinct factorizations g = x1 xn = y1 yn , and let i be the smallest index such that xi = yi . Then x1 = y1 , . . . , xi-1 = yi-1 , and cancellation yields xi xi+1 xn -1 yi xi = = yi yi+1 yn yi+1 yn x-1 x-1 n i+1
The right hand product lies in the subgroup Ai+1 An Hi , while the left hand product lies in Ai . We are assuming the intersection Ai Hi is trivial, so both products reduce to the identity element e and we get xi = yi , contrary to the definition of i. Uniqueness of factorization is proved. We are now ready to state the main result. We frame it in terms of the unique factorization condition (iii), but by 6.1.11 we may replace (iii) with (iii) if we wish. 6.1.12 Theorem (Internal Direct Product). Let G be a group and A1 , . . . , An subgroups such that (i) Each Ai is a normal subgroup in G (ii) The product set A1 An is equal to G. (iii) Each g G decomposes uniquely as g = a1 an with ai Ai . Then G is isomorphic to the direct product group A1 . . . An . In particular, elements of Ai and Aj automatically commute if i = j. Proof: If i = j we have pairwise disjointness Ai Aj = (e) because Aj Hj , and Ai Hj = (e) by 6.1.11. Using this we can show that Ai and Aj commute when i = j. -1 To see why, let xi Ai , yj Aj and consider the "commutator" z = xi yj x-1 yj . Since i -1 -1 Aj is normal in G the element z = (xi yj xi )yj must lie in Aj ; the same argument shows that z also lies in Ai . Since i = j, we get z = e by unique factorization. But -1 xi yj x-1 yj = e implies xi yj = yj xi , as required. i The unique decomposition property (iii) means precisely that the map p(a1 , . . . , an ) = a1 . . . an from the Cartesian product set A1 . . . An to G is a bijection. It is also a homomorphism, because if a = (a1 , . . . , an ), b = (b1 , . . . , bn ) we can make the following rearrangements of commuting group elements p(a)p(b) = = = a1 an b 1 b n a1 b 1 a2 an b 2 b n a1 b 1 a2 b 2 a3 an b 3 b n . . . a1 b 1 a2 b 2 . . . an b n p(ab) (since ab = (a1 b1 , . . . , an bn ))
= =
Thus p is an isomorphism of groups. That finishes the proof. In view of this result, a group satisfying the conditions of Theorem 6.1.12 is often called an internal direct product of the subgroups A1 , . . . , An . This decomposition theorem, and the notion of direct product, are most often applied when there are just two factors. Then the decomposability criteria are much simpler, and direct products structure G A B is much easier to recognize. = 6.1.13 Corollary. Let G be a group and A, B two subgroups such that (i) A and B are normal subgroups in G. (ii) The product set AB is equal to G. (iii) A B = (e). 4
Then G is isomorphic to the direct product A B under the map p(a, b) = a b. Proof: Every g G has a factorization g = ab by (ii). This is unique by (iii) because ab = a b (a )-1 a = b(b )-1 The expression on the left is in A while the one on the right is in B, so by (iii) both expressions are equal to e; that means a = a and b = b as required for uniqueness. Unique factorization determines a bijection f : G A B if we take f (g) = f (ab) = (a, b). To see that f is a group isomorphism when the Cartesian product set is given the direct product structure (1) we first note that elements of A commute with elements in B. This holds if all "commutators" [a, b] = aba-1 b-1 are trivial. But the commutator can be written two ways. We see that [a, b] is in B because [a, b] = (aba-1 ) b = b b and B is normal. If instead we write [a, b] = a(ba-1 b-1 ) = aa we conclude that the commutator is also in A because A is normal. By (iii) the commutator must be trivial. From commutativity ab = ba and the definition of products in A B we get f (ab a b ) = f (aa bb ) = (aa , bb ) = (a, b) (a , b ) = f (ab) f (a b ) so f : G A B is an isomorphism. The following examples illustrate the use of these results. 6.1.14 Example. The cyclic group G = Z4 , with generator a = [1], is an abelian group of order 4. So is the direct product G = Z2 Z2 , whose elements can be written as e = (0, 0) a = (1, 0) b = (0, 1) c = (1, 1)
where by abuse of notation we write k for the class [k]. Can these groups be isomorphic i.e. is the cyclic group a direct product of smaller subgroups? Answer: No. In G every element g = e has order order o(g) = 2; in fact, writing (+) for the operation in Z2 Z2 we have (a, b) + (a, b) = (2 a, 2 b) = (0, 0). In contrast, G has a cyclic generator a such that o(a) = 4. That's impossible if G G . = 6.1.15 Example. Let G be the cyclic group (Z6 , +). Since the order of any element in G must divide |G| = 6, the possible orders are o(g) = 1, 2, 3, 6. The element a = [3] has order 2, since [3]+[3] = [6] = [0], and the group A it generates has order |A| = 2; obviously A (Z2 , +) because, up to isomorphism, Z2 is the only group of order 2. Similarly, the = element b = [2] has additive order o(b) = 3, and generates a cyclic subgroup of order |B| = 3. Obviously B (Z3 , +). = G is abelian so both A and B are normal subgroups. Furthermore, AB is a subgroup of both A and B, so by Lagrange its order must divide both |A| = 2 and |B| = 3; that forces A B to be trivial. Finally, the product set A + B is easily seen to be all of G. (We are writing the group operation as (+) in this example.) By 6.1.13 we conclude that Z6 Z2 Z3 . This cyclic group (Z6 , +) does decompose as the direct product of smaller = cyclic subgroups. What might account for the different outcomes in the last two examples? A complete answer will emerge eventually when we discuss the Chinese Remainder Theorem (below), but for the moment it suffices to point out that the subgroups A, B in the second example are associated with different prime divisors of |G| = 6. There are no distinct prime divisors when |G| = 4 = 22 .
5
6.1.15A Exercise. Show that the following map : Z6 Z3 Z2 ([i]6 ) = ([i]3 , [i]2 ) has the properties (a) is well defined (independent of choice of class representative i). (b) is a bijection. (c) is a homomorphism between (Z6 , +) and the direct product group Z3 Z2 . so that is an isomorphism. 6.1.15B Exercise. In 6.1.15 we showed Z3 Z2 Z6 . Is Zm Zn always isomorphic = to Zmn ? Prove or give a counterexample. 6.1.16 Exercise. Do you think it possible to decompose the cyclic group (Z15 , +) as a direct product of smaller cyclic subgroups? How about the cyclic group (Z9 , +)? Explain. Hint: If G factors, what orders could the factors have? 6.1.17 Example. Up to isomorphism, describe all groups of order |G| = 4. Discussion: We procede by looking at the largest possible order o(b) for an element of G. By Lagrange, the only possibilities are o(b) = 1, 2, 4; since o(b) = 1 b = e the maximal order cannot be 1. Case 1: o(b) = 4. Then G is cyclic of order 4 and must be isomorphic to (Z4 , +). Case 2: o(b) = 2. Then all elements x = e have order o(x) = 2, so that x2 = e and x-1 = x for each x G. Pick any a = e and let A = a ; obviously A Z2 . Next = pick any element b A and let B = b ; obviously B Z2 too, but B = A. The / = subgroup A B lies within both A and B, and must be trivial; otherwise |A B| = 2 and A = B = A B, which has been excluded by our choice of b. We claim that the product set AB is equal to G. A sophisticated way to verify this is to invoke Theorem 3.4.7 which says that |AB| = |A| |B| / |A B| = (2 2)/1 = 4, and hence that AB = G. Finally we observe that G must be abelian. In fact if a, b G the element ab has e = (ab)2 = abab; if we then multiply on the right by b-1 and on the left by a-1 we get a-1 b-1 = ba. But a-1 = a and b-1 = b, so ab = ba as required. It follows that A and B are normal subgroups in G, and hence by 6.1.14 that G is the internal direct product of these subgroups: G Z2 Z2 in Case 2. = Thus, up to isomorphism, the only possibilities for a group of order four are the groups G Z4 and G Z2 Z2 of Example 6.1.14. = = Here is an example in which previous results are used to exhibit the presence of direct product structure in a group. It follows the lines of 6.1.17, except for the need to invoke the Cauchy theorems (Cor. 4.3.3) to establish commutativity of the group. *6.1.18 Exercise (Groups of order p2 ). Assume G is a finite group of order |G| = p2 for some prime p > 1. Prove that G Zp2 or G Zp Zp . = = Hint: Use the Cauchy theorems and adapt the ideas developed in 6.1.17. 6.1.19 Exercise. Prove that |A1 . . . Ar | = |A1 | . . . |Ar | for any direct product of groups. *6.1.20 Exercise. Prove that the order o(x) of an element x = (a, b) in the direct product group A B is the least common multiple lcm(o(a), o(b)). Is a similar result true for direct products of several groups? Hint: Observe that xm = e am = e and bm = e. In any group, g m = e m is a multiple of o(g). 6 ([i]6 Z6 )
6.1.21 Exercise. Is Z15 Z4 a cyclic group i.e. does it have an element of order 60? Does Z15 Z5 have elements of order (i) 75? (ii) 25? (iii) 15? (iv) 3? (v) any other order? 6.1.22 Exercise. Identify all elements x = (a, b) of order o(x) = 5 in Z15 Z5 . Same for Z15 Z4 . Ultimately we will have a lot to say about the structure of a finite group G in terms of the prime divisors of its order n = |G|. There is one important case in which the outcome has a striking simplicity. 6.1.23 Theorem. Let G be a nontrivial finite group of prime order |G| = p > 1. Then G is isomorphic to the cyclic group (Zp , +). Furthermore, every b = e is a cyclic generator. Proof: Let b = e and consider the cyclic subgroup H = {e, b, b2, . . . , bk-1 }, with bk = e. By Lagrange, |H| = k 2 must divide the order |G| = p of the whole group. Hence k = p and H = G. Direct products and the Chinese Remainder Theorem. The Chinese Remainder Theorem (CRT) has its roots in number theory but has many uses. One application completely resolves the issues regarding direct products Zm Zn mentioned in 6.1.14 16. The original remainder theorem arose in antiquity when attempts were made to solve systems of congruences involving several different moduli ni x a1 . . . x ar (mod n1 ) xZ (mod nr )
The notion of congruence is a modern one; the ancient chinese would have viewed this problem as the search for an integer x with specified remainders ai after division by ni , i = 1, 2, . . . r. Such systems do not always have solutions, but solutions do exist if the moduli n1 , . . . , nr are pairwise relatively prime, so that gcd(ni , nj ) = 1 if i = j, and then the solution x is unique up to added multiples of the least common multiple m = lcm(n1 , . . . , nr ) of the moduli. 6.1.24 Example. Here are two systems of congruences (a) x5 x1 (mod 3) (mod 12) (b) x5 x1 (mod 3) (mod 5)
Taking a "bare hands" approach we shall verify that the system (a) has no solutions and that the solutions of (b) are of the form x0 + k 15 where x0 = 11 and k Z. Discussion: If x Z solves (a) there must exist k, Z such that x = 5 + 3k = 1 + 12 But then k and satisfy the equation 12 = 4 + 3k, which would imply 4 = 12 - 3k is in Z 12 + 3 Z = gcd(12, 3) Z = 3Z That's impossible because 4 is not a multiple of 3. For a solution x of (b) we would have x = 5 + 3k = 1 + 5 as above, which means that k, satisfy 4 = -3k + 5. Since gcd(3, 5) = 1 there exist integers r, s such that 3r + 5s = 1; multiplying through by 4 we get 3(4r) + 5(4s) = 4 7
which tells us how to find k and such that 4 = -3k + 5; just take k = -4r and = 4s. Then take x = 5 + 3k or x = 1 + 5 to get a solution x Z of the congruance (a). (You can check that both choices give the same value of x.) As for uniqueness, if x is any other solution then y = x - x 0 (mod 3) and also (mod 5), so y is a multiple of 3 and 5, and of lcm(3, 5) = 15. On the other hand it is obvious that any x x + 15Z is a solution, so solutions are unique (mod 15). Note that solution of the problem rests on our ability to compute c = gcd(a, b) efficiently via the GCD Algorithm, and to write c in the form c = ra+ bs by working that algorithm backwards. 6.1.24A Exercise. Solve the following congruences by hand or determine that they have no solutions. (a) x 2 (mod 15) x 5 (mod 7) (b) x3 x5 (mod 9) (mod 12) (c) x2 x5 (mod 15) (mod 19)
6.1.25 Exercise. Recall that the greatest common divisor gcd(a1 , . . . , ar ) of a1 , . . . , ar is the smallest positive element in the lattice = Za1 + . . . + Zar of integer-linear combinations. (The proof is the same as when r = 2.) We say that the ai are jointly relatively prime if this gcd = 1. Prove that (a) (Pairwise relatively prime) (Jointly relatively prime) (b) Give an example showing that the converse does not hold. To keep things simple let's consider a system involving just two congruences (5) x a1 (mod m) x a2 (mod n)
This remainder problem is completely equivalent to a problem in group theory: the system is solvable for every choice of a1 , a2 on the right if and only if the direct product group Zm Zn is cyclic i.e. if and only Zm Zn Zmn . = 6.1.26 Theorem (Chinese Remainder Theorem). If m, n > 1 are relatively prime, so that gcd(m, n) = 1, then Zm Zn Zmn as additive groups. Furthermore, the system = of congruences (5) has a solution for every choice of a1 , a2 Z, and if x0 Z is one solution the full set of solutions in Z is the congruence class x0 + Zmn. Proof: To keep track of the three different types of congruence classes we write [k]m = k + Zm, [k]n = k + Zn, [k]mn = k + Zmn to distinguish them. Now observe that any element [k]mn in Zmn determines well-defined classes [k]m , [k]n in Zm , Zn having the same representative k because the correspondences [k]mn [k]m and [k]mn [k]n
are well defined maps from Zmn into Zm and Zn respectively. In fact, if k is any other representative of the class [k]mn , that means k = k + s mn for some integer s. But k = k + (sn)m is congruent (mod m) to k and hence [k ]m = [k]m ; likewise [k ]n = [k]n . We now create a map from Zmn to the Cartesian product set Zm Zn by setting (6) ([k]mn ) = ([k]m , [k]n ) Zm Zn for all [k]mn in Zmn
This map is well-defined independent of how we choose representatives k of elements of Zmn . It is immediate from the definition that intertwines the (+) operations in the
8
additive groups Zmn and Zm Zn because ([k]mn + []mn ) = = = = ([k + ]mn ) = ([k + ]m , [k + ]n ) ([k]m + []m , [k]n + []n ) ([k]m , [k]n ) + ([]m , []n ) ([k]mn ) + ([]mn ) (defn. of (+) in Zm Zn )
Thus : (Zmn , +) (Zm , +) (Zn , +) is a homomorphism of abelian groups. So far we have not used the hypothesis gcd(m, n) = 1, which we invoke to show that is one-to-one. (It will then be surjective too, because the sets Zmn and Zm Zn have the same cardinality mn.) If two points in Zmn have the same image, say with ([k]mn ) = ([]mn ), the components in Zm Zn must match up so that (7) [k]m = []m [k]n = []n which means that k - is divisible by m k - is divisible by n
We need to show that k (mod mn) or k - 0 (mod mn). In general, divisibility of an integer r by m and n is not enough to insure that r is divisible by mn (try m = 4, n = 2, r = 8). But if m and n are relatively prime they have no prime factors in common. By comparing the prime factorizations of m, n and k - we see that (m and n both divide k - ) = (mn divides k - ) Thus [k]mn = []mn , as required to show that is one-to-one. The map is the desired isomorphism between groups. To connect all this with the remainder problem: If m, n are relatively prime the element 1 = ([1]m , [1]n ) must be a cyclic generator for Zm Zn because it is the image of the generator [1]mn in the cyclic group Zmn . Given a1 , a2 , consider the element a = ([a1 ]m , [a2 ]n ) in the product group. Then some multiple of 1 is equal to a, say a = k 1: ([a1 ]m , [a2 ]n ) = a = k1 = (k[1]m , k[1]n ) = ([k]m , [k]n ) in Zm Zn
so that k a1 (mod m) and k a2 (mod n). That makes x0 = k a solution of the congruences (5). If k is another solution then k 1 = a = k 1 so we get (k - k) 1 = a - a = 0 = ([0]m , [0]n ). But 1 has order mn, being the generator of Zmn , so this happens (k - k) is a multiple of mn. The full set of solutions is therefore x0 + Z mn as claimed. 6.1.27 Exercise. Is Z45 Z9 Z5 ? Is Z45 Z15 Z3 ? Is Z3 Z4 Z5 Z60 ? = = = Hint: For the last part try writing Z3 Z4 Z5 = (Z3 Z4 ) Z5 . *6.1.28 Exercise. In Z7 Z12 is there some "additive exponent" k such that k 1 = k ([1]7 , [1]12 ) is equal to ([3]7 , [-4]12 ) ?
If so, find a k that does this. Then find a "normalized " solution lying in the range 0 k < 7 12 = 84. The underlying idea in proving 6.1.26 is illustrated in Figure 6.1 where we represent Zm Zn as an m n array of squares. Counting from the bottom left square 0 = (0, 0), the element 1 = (1, 1) is the dark shaded square and its additive "powers" 1, 1 + 1 = (2, 2), . . . , k1 = (k, k), . . . move diagonally upward until they hit an edge, at which point they re-enter the array from the opposite edge. If gcd(m, n) = 1 all squares get hit exactly once until we finally 9
Figure 6.1 The product group Zm Zn represented as an m n array. The element 1 = (1, 1) is shaded and its iterates are marked "x". In this portrayal the array is 7 12; since gcd(7, 12) = 1 the iterates will cycle once through each point in the array before returning to position 1.
arrive back at 0 = mn 1; the next step takes us to 1 = (1, 1) and then the pattern repeats. This observation tells us how to compute the inverse -1 : Zm Zn Zmn of the isomorphism constructed in proving 6.1.26. A pair a = ([a1 ]m , [a2 ]n ) has preimage [k]mn back in Zmn k is the first integer k = 0, 1, 2, . . . such that k 1 = a. This k is of course the desired solution of the congruence system (5), so knowing how to compute -1 is equivalent to being able to solve the remainder problem. Systematic Solution of Remainder Problems. When gcd(m, n) = 1 we know that solutions exist and are unique up to an added multiple of mn; the problem is to lay hands on an actual particular solution x0 , say one normalized so that 0 x0 < mn. Let's work backward: if x0 is a solution then [x0 ]m = [a1 ]m [x0 ]n = [a2 ]n = there exist R, S Z such that x0 = a1 + Rm x0 = a2 + Sn
Thus x0 = Rm + a1 = Sn + a2 which implies a2 - a1 = Rm - Sn. Since gcd(m, n) = 1 we can find r, s Z such that 1 = rm + sn = rm - (-s)n. Multiplying both sides by (a2 - a1 ) yields (a2 - a1 ) = [(a2 - a1 )r] m - [(a2 - a1 )(-s)] n We obtain the following values for R and S R = (a2 - a1 )r S = -(a2 - a1 )s
A particular solution is then x0 = Rm + a1 = Sn + a2 (both values being equal). 6.1.29 Exercise. Use the method outlined above to find the solutions of the congruence problem x 14 (mod 18) x -2 (mod 25) Find a particular solution lying in the range 0 x0 < 480 = lcm(18, 25). 6.1.29A Exercise. If gcd(m, n) = c > 1 can we ever have Zm Zn Zmn ? = The CRT can be generalized in many ways. It can be made to work, with essentially the same proof, for congruence systems of arbitrary size provided we require that the moduli n1 , . . . , nr be pairwise relatively prime, so that gcd(ni , nj ) = 1 if i = j. This is a much stronger condition than saying gcd(n1 , . . . , nr ) = 1, which means that no prime p > 1 divides every ni . 6.1.30 Exercise. Using induction on r prove that Zn1 n2 ...nr = Zn1 . . . Znr if the ni are pairwise relatively prime (so gcd(ni , nj ) = 1 if i = j). 6.1.31 Exercise. Give an example of integers n1 , n2 , n3 > 1 that are jointly relatively 10
prime (so that gcd(n1 , n2 , n3 ) = 1: there is no single prime p such that p|ni for all i) but the ni are not pairwise relatively prime (so gcd(ni , nj ) = 1 and ni , nj have no primes in common if i = j). In another direction, we observe that Zn comes equipped with both a (+) and a multiplication operation (), making it a commutative ring with identity as in Chapter 2. One can define a direct product R1 R2 of commutative rings very much as we defined direct product of groups, by imposing the following sum and product operations on the Cartesian product set R1 R2 = {(a, b) : a R1 , b R2 }
(r1 , r2 ) + (r1 , r2 ) = (r1 , r2 ) (r1 , r2 ) = (r1 + r1 , r2 + r2 ) (r1 r1 , r2 r2 )
It is easy to check that (R1 R2 , +, ) is a new commutative ring, with identity element 1 = (11 , 12 ) if each Rk has an identity element 1k . A review of the proof of the Chinese Remainder Theorem reveals that the bijective map : Zmn Zm Zn created there is actually an isomorphism of commutative rings because it intertwines both the (+) and () operations ([k]mn + []mn ) ([k]mn []mn ) = ([k]mn ) + ([]mn ) (proved in 6.1.26)
= ([k]mn ) = ([k]m , [k]n ) = ([k]m , [k]n ) ([]m , []n ) = ([k]mn ) ([]mn ) (defn. of () in Zm Zn )
Thus Zmn Zm Zn as rings, not just as groups. One consequence is a result about = groups of units in Zn that has important consequences in number theory, although we will just view it as a result about the direct product structure of certain groups of units. 6.1.32 Theorem. If m, n > 1 are relatively prime and Um , Un , Umn are the multiplicative abelian groups of units in Zm , Zm , Zmn then (8) (Umn , ) (Um , ) (Un , ) =
and in particular the sizes of these groups satisfy the following multiplicative condition (9) |Umn | = |Um | |Un | if gcd(m, n) = 1
Before launching into the proof we remark that the set of units in any commutative ring R with identity 1R R is defined just as for Zn (10) UR = {x R : y R such that x y = 1R }
The element y in (10) is called the multiplicative inverse of x, written y = x-1 . As with Zn , the units UR form a commutative group (UR , ) under multiplication. The proof of 6.1.32 rests on two easily verified properties of groups UR . *6.1.33 Exercise. If : R R is an isomorphism of commutative rings with identity show that (a) maps identity element to identity element: (1R ) = 1R . (b) maps units to units, (UR ) = UR , and is an isomorphism of these multiplicative groups. 11
6.1.34 Exercise. If R1 , R2 are commutative rings with identities 11 R1 , 12 R2 show that the set of units UR1 R2 in the direct product ring is the Cartesian product of the separate groups of units (11) UR1 R2 = UR1 UR2 = {(x, y) : x UR1 , y UR2 }
Proof (6.1.32): First observe that : Zmn Zm Zn maps Umn one-to-one onto the group of units UZm Zn , by 6.1.33. By 6.1.34, the group of units in Zm Zn is equal to Um Un . Since is a bijection that intertwines the () operations on each side it is a group isomorphism from (Umn , ) to the direct product group (Um , ) (Un , ). Remarks: Even if m and n are relatively prime, the orders |Um |, |Un | of the groups of units need not have this property. For instance, if p > 1 is a prime then |Up | = p - 1 and there is no simple connection between p and the prime divisors of p - 1, or between the divisors of |Up | = p - 1 and |Uq | = q - 1 if p, q are different primes. Furthermore the groups Um , Un need not be cyclic, though they are abelian. Nevertheless we get a direct product decomposition (8). Theorem 6.1.32 operates in a very different environment from that of the Chinese Remainder Theorem 6.1.26. Its proof is also more subtle in that it rests on the interplay between the (+) and () operations in Z, while the CRT spoke only of (+). 6.1.35 Exercise. If n1 , . . . , nr are pairwise relatively prime, with gcd(ni , nj ) = 1 for i = j, use induction on r to prove that Un1 n2 ...nr Un1 . . . Unr = and |Un1 n2 ...nr | = |Un1 | . . . |Unr |
. 6.1.36 Exercise. Write Z630 as a direct product Zn1 . . . Znr of cyclic groups. Using 2.5.30 we could compute |U630 | by tediously comparing the prime divisors of 630 = 2 32 5 7 with those of each 1 k < 630. However, the formula of 6.1.35 might provide an easier way to calculate |U630 |. Do it. Hint: Is the group of units (U9 , ) cyclic? (Check orders o(x) of its elements.) The Euler Phi-Function : N N is often mentioned in number theory. It has many equivalent definitions, but for us it is easiest to take (1) = 1 (n) = |Un | = #(multiplicative units in Zn )
Theorem 6.1.32 shows that (mn) = (m)(n) if m and n are relatively prime, and it is this property that makes the -function so useful. *6.1.37 Exercise. The multiplicative property of Exercise 6.1.32 is no help in computing the size of the group of units Upk where p is a prime, since pk does not split into relatively prime factors. (Think p = 2, k = 7, pk = 128.) However, one can show directly that Upk = #{1 j < pk : gcd(j, pk ) = 1} = #{1 j < pk : p does not divide j} has cardinality |Upk | = pk-1 (p - 1). Carry out this computation. Hint: Every integer 0 m < pk has a unique "base p" expansion m = a0 + a1 p + a2 p2 + . . . + ak-1 pk-1 where 0 aj < p for each j. Some divisiblity questions become easy using base p expansions. You may want to use the following exercise too. 6.1.38 Exercise. Prove that gcd(m, pk ) = 1 m is a multiple of p. 6.1.39 Exercise. Compute the size |Un | for n = 2 32 53 49 = 110, 250. 12
By exploiting the multiplicative property of the -function we obtain a group-theoretic proof of the following important fact from number theory that reflects the subtle interplay between the operations (+) and () in the ring Zn . 6.1.40 Theorem. For any integer n 1 we have n=
d|n,1dn
(d) =
d|n,1dn
|Ud |
Proof: In the cyclic group (Zn , +) every element x has some additive order o+ (x) = smallest integer k such that k x = [0]; by Lagrange, this must be a divisor of n. Letting Sd = {x Zn : o+ (x) = d} we obviously have Zn = d|n Sd . These sets are disjoint, and they are all nonempty because, as in Theorem 3.4.7, there is a unique cyclic subgroup Hd of order d in Zn for each divisor d|n. Its generator lies in Sd so Sd is nonempty, but by uniqueness of the group Hd we must have x = Hd for each x Sd , so that Sd Hd . Under the isomorphism Hd (Zd , +) the set Sd corresponds to the cyclic generators in = (Zd , +). But in Proposition 3.1.33 we identified these generators explicitly as the set of units Ud Zd . It follows that n=
d|n
|Sd | =
d|n
|Ud | =
d|n
(d)
as claimed.
6.2 Semidirect products.
We have determined all groups of order 1 |G| 5 using the notion of direct product. But when |G| = 6 we encounter some cases that cannot be understood in terms of cyclic groups and their direct products. This leads us to a more general product construc| tion, the semidirect product N H of two groups. The construction is motivated by the following considerations. If A and B are subgroups of a group G, the conditions (12) AB = G and A B = (e)
imply that every g G has a unique factorization g = ab. Existence is clear since G = AB, while uniqueness follows from the condition A B = (e) because a1 b 1 = a2 b 2 (a2 )-1 a1 = b2 (b1 )-1 is in A B (a2 )-1 a1 = e and b2 (b1 )-1 = e a2 = a1 and b2 = b1 This means there is a natural "parametrization" of points in G by pairs (a, b) in the Cartesian product space A B, which at the moment is not equipped with a group structure. The correspondence A B G is effected by the bijection p : A B G with p(a, b) = a b (product of a and b in G). The labels a and b may be thought of as "coordinates" labeling all points in G, and it is natural to ask what the group operation looks like when described in terms of these parameters. If both A and B are normal subgroups then by 6.1.13 G is their internal direct product and our question has a simple answer. When we identify G AB the group operations take the familiar form (a, b) (a , b ) = (aa , bb ) and the identity element is e = (e, e). 13 (a, b)-1 = (a-1 , b-1 )
If neither subgroup is normal, the description of the group law in terms of these parameters can be formidable, and we will not attempt to analyze this situation here. There is, however, a fruitful middle ground: the case in which just one of the subgroups is normal. This leads to the notion of semidirect product, which encompasses an enormous set of examples that arise in geometry and higher algebra. In this setting we relabel the subgroups as N and H with N the normal subgroup. That way it will be readily apparent in calculations which group elements lie in the normal subgroup. Points in G correspond to pairs (n, h) in the Cartesian product set N H. Furthermore, the product operation in G can be described explicitly in terms of these pairs. Given elements g1 = n1 h1 , g2 = n2 h2 in G we must rewrite g1 g2 = n1 h1 n2 h2 as a product n h with n N, h H. The problem is, essentially, to move h1 to the other side of n2 , keeping track of the damage if they fail to commute. This is easy. Because N G we may multiply and divide by h1 to get n1 h1 n2 h2 = n1 (h1 xh-1 ) h1 h2 = n h 1 in which n N because h1 N h-1 N . Identifying each gi with its corresponding 1 coordinate pair (ni , hi ) in N H, the group multiplication law takes the form (n1 , h1 ) (n2 , h2 ) (13)
-1 = (n1 (h1 n2 h1 ) , h1 h2 )
= (n1 h1 (n2 ), h1 h2 )
where h : N N is the conjugation operation h (n) = hnh-1 for any h H. The normal subgroup N is invariant under the action of any conjugation g (n) = gng -1 , g G, and h is just the restriction h = h |N of the inner automorphism h to the set N . Obviously h Aut(N ) for each h H, and we obtain a group action H N N of H on N such that e = idN h1 h2 = h1 h2 h-1 = (h )-1 (inverse of the operator h )
which means that the correspondence : H Aut(N ) given by (h) = h is a group homomorphism. Conversely, as we noted in our previous discussion of group actions (Chapter 4), every such homomorphism determines an action of H on N by automorphisms. There is a one-to-one correspondence between actions H N N and homomorphisms : H Aut(N ). It is evident from (13) that the original group G can be reconstructed from its subgroups N, H if we know the action of H on N via conjugation operators h . In this way G has been realized as an "internal semidirect product" of N by H i.e. as the Cartesian product set N H equipped with the multiplication law (13) determined by h . We summarize these remarks as follows. 6.2.1 Proposition (Internal Semidirect Product). Let G be a group and N, H two subgroups such that (i) N is normal in G (ii) N H = G (iii) N H = (e)
Each g G has a unique factorization as g = nh, so there is a natural bijection between G and the Cartesian product set N H. The group action H N N via conjugation operators h (n) = hnh-1 determines the following "product operation" in the Cartesian product set N H. (14) (n, h) (n , h ) = (nh (n ), hh ) for n, n N and h, h H
This operation makes the Cartesian product set into a group, and the product map p(n, h) = nh is an isomorphism from (N H, ) to the original group G. 14
Proof: We have noted the unique factorization g = nh in the remarks following (12); the bijection N H G is given by the product map p(n, h) = nh. Our other claims follow if we can just prove that the product map p intertwines group operations in N H and G: p((n, h) (n , h )) = p(n, h)p(n , h ) From definition (14) we see that p(n, h)p(n , h ) = nhn h = n(hn h-1 )hh = nh (n )hh = p(nh (n ), hh ) = p((n, h)(n , h )) (last step by definition (14)). All this is expressed by saying that G is the internal semidirect product of the | subgroups N and H, which we indicate by writing G = N H to distinguish it from the direct product N H of Section 6.1. Of course if the action H N N is trivial, as when the subgroups H and N commute (so that h = idN for all h H), then the semidirect product reduces to the ordinary direct product.
| 6.2.2 Exercise. In a semidirect product G = N H with group law (14) it is clear that the identity element is just the pair e = (e, e). Use this observation to compute the inverse
(15)
(n, h)-1 = (h-1 n-1 h, h-1 ) = (h-1 (n-1 ), h-1 )
for all n N, h H
for any pair (n, h) N H.
| 6.2.3 Exercise. In a semidirect product G = N H with group law (14) the pairs (n, e) and (e, h) in the product set N H correspond under p to the group elements n and h. Use (14) and (15) to verify that
(a) (n, e) (e, h) = (n, h) (b) Conjugation of (n, e) by (e, h) under the group law (14) has the same effect | in N H as conjugating n by h in the original group G i.e. we have (e, h)(n, e)(e, h)-1 = (hnh-1 , e) = (h (n), e) What happens if we multiply (e, h) (n, e), reversing the order of the elements in (a)?
| *6.2.3A Exercise. In a semidirect product N H prove that conjugating an element x = (n , h ) by an element g = (n, h) yields
g (x) = (n, h)(n , h )(n, h)-1 = (nh (n )h h-1 (n-1 ) , hh h-1 ) Write out the simplified formula for the action of an arbitrary g = (n, h) in G on an element x = (n , e) of the normal subgroup N . The following observation often simplifies the job of deciding whether a group G is in fact a semidirect product of two of its subgroups. 6.2.4 Exercise. Let G be a group and N, H subgroups such that (i) N H = G, (ii) N H = (e), and (iii) H normalizes N in the sense that hN h-1 N for every h H.
| Prove that N is in fact normal in G, so G is the internal semidirect product N H.
External Semidirect Product. We now reverse this bottom-up analysis, in which N
15
| and H lie within a pre-existing group G, to construct a new group N H, the external semidirect product of N by H, given the following ingredients
(i) Two unrelated abstract groups N and H (16) (ii) A group action H N N implemented by automorphisms h Aut(N ) for each h.
Of course, specifying the action in (ii) is completely equivalent to specifying some homomorphism : H Aut(N ); just set (h) = h . 6.2.5 Theorem. Given abstract groups N, H and a homomorphism : H Aut(N ) define a binary operation on the Cartesian product set N H, exactly as in (14) and (15). (17) (n, h) (n , h ) = (nh (n ), hh ) for all n, n N and h, h H
where h = (h) Aut(N ). Then G is a group, the external semidirect product, | which we denote by N H. The inversion operation g g -1 in this group has the form (15). The subsets N = {(n, e) : n N } and H = {(e, h) : h H} are subgroups in G that are isomorphic to N and H. They satisfy the conditions N H = G, N H = (e), and N is normal in G. Thus G is the internal semidirect product of these subgroups. Proof: There is some bother involved in checking that the operation (17) is associative because the action does not arise via multiplications in some pre-existing group, as it did in 6.2.1. We leave this routine but tedious calculation to the reader. The identity element is obviously e = (e, e) = (eN , eH ). The inverse operation is (n, h)-1 = (h-1 (n-1 ), h-1 ) In fact, recalling that e = idN and that h-1 = (h )-1 because : H Aut(N ) is a homomorphism, (17) gives
-1 (nh (h (n-1 )), hh-1 ) = (nn-1 , e) = (e, e) (h-1 (n-1 ), h-1 ) (n, h) = (h-1 (n-1 )-1 (n), h-1 h) h -1 -1 -1 = ( (h (n)) h (n), e) = (e, e)
(n, h) (h-1 (n-1 ), h-1 ) =
In G the maps n (n, e) N and h (e, h) H are isomorphic embeddings of N and H in G because they are bijections such that (n, e) (n , e) = (nn , e) and (e, h) (e, h ) = (e, hh )
We get G = N H because (n, e) (e, h) = (ne (e), eh) = (n, h), and it is obvious that N H = {(e, e)}. Normality of N follows because (n, h)(n , e)(n, h)-1 = (n, e)(e, h)(n , e)(e, h-1 )(n-1 , e) = (n, e)(h (n ), h)(e, h-1 )(n-1 , e) = (n, e)(h (n ), e)(n-1 , e) = (nh (n )n-1 , e) for all n N ; the lefthand component is back in N so that gNg -1 N for any g = (n, h) G. When we identify N N and H H, the action of h H by conjugation on n N = = matches up with the original action of H N N determined by , so the group G we | have constructed is the internal semidirect product N H of these subgroups. 16
Figure 6.2. The basic symmetry operations on a regular n-gon, shown here for the regular hexagon (n = 6). The same idea works for all n. In our discussions vertices are labeled 0, 1, 2, . . . , n - 1 as indicated. The symmetry groups Dn have different properties for even and odd n, and n = 2 is exceptional.
The next example is important in geometry. 6.2.6 Example (The Dihedral Groups Dn ). These groups of order |Dn | = 2n, defined for n 2, are the full symmetry groups of regular n-gons. To describe Dn consider a regular n-gon in the xy-plane, centered at the origin and with one vertex on the positive x-axis, as shown in Figure 6.2 (where n = 6). Let = 2/n radians, and define the basic symmetry operations = (counterclockwise rotation about the origin by radians) = (reflection across the x-axis)
Obviously these elements have orders (18A) o() = 2 so 2 = e and -1 = o( ) = n with distinct powers j for 0 j < n and n = I.
The dihedral group Dn is the subgroup Dn = , generated by and in the group O(2) of linear isometries (distance-preserving transformations) of the plane. Obviously N = is a cyclic subgroup isomorphic to Zn and H = is a copy of Z2 embedded | in Dn . We will show that N is normal and Dn is the semidirect product N H. We begin by verifying another relation, which together with the obvious relations (18A) completely determines Dn . (18B) = -1 = -1 (Note that -1 = - and -1 = ) This relation tells us how to pass a rotation across a reflection when forming products, since = -1 ; furthermore, since conjugation by is an automorphism we also get k = k -1 = ( -1 )k = -k for all k Z
One could check (18B) by tedious matrix computations, but it is easier simply to track the action of each factor on the standard unit vectors e1 = (1, 0), e2 = (0, 1) in R2 , as shown in Figure 6.3 below. Once we know what the product does to basis vectors we
17
Figure 6.3. Action of on basis vectors in R2 .
know what it does to all vectors, and it is not hard to recognize the outcome as a familiar linear operator. (In (18B) the end result is clockwise rotation by radians.) Note also what happens if we repeatedly conjugate an element k N by the reflection operation . (18C) i k -i = i k i =
(-1)i k
The reason is simple: conjugation by yields -k , doing it again reverses the sign on the exponent one more time, leaving k . Performing this "sign flip" i times yields
(-1)i k
=
k -k
if i is even if i is odd
Since and generate Dn the identities (18A) and (18B) imply that N = is a nomal subgroup in Dn , as a consequence of the following general result. 6.2.6A Exercise. If elements a1 , . . . , an generate a group G, prove that a subgroup H is normal in G H is invariant under congugation by the generators: ai Ha-1 H i for all 1 i n
Note: In practice it is much easier to verify normality by checking the action of a few generators, than to examine the action of an arbitrary group element. In the last example there were just two generators. Using the relations (18) we now show that every element of Dn = , can be written uniquely in the form k where 0 k < n and = 0 or 1
Since n = 2 = I, the operator k depends only on the (mod n) congruence class of k, and it is convenient to think of this exponent as an element k Zn ; likewise the exponent in can be thought of as an element Z2 . Then the factorization of elements in Dn takes the form (19) Any element in the dihedral group Dn can be written uniquely in the form k where k Zn and Z2 .
To prove this, we show that the set S of elements listed in (19) is already a group. Then, since Dn = , S and S is a subgroup containing the generators, the two sets must
18
be equal, so Dn = S as claimed. Clearly I S, and S S S because (k )(r s ) = k r ( - ) s = k ( r - ) +s k - r +s = ( ) (conjugation by is an automorphism) = k =
(-1) r
+s
(by (18C))
k+(-1) r
+s S
Finally S -1 = S, and S is a subgroup, because (k )-1 = = = = = - -k -k - ( )-k -
(-1)
(because = -1 )
-k
- S
(by (18C))
-(-1) k -
From (19) we see that Dn is a finite group of operators on the plane, with |Dn | = 2n. Summarizing these observations we have 6.2.7 Corollary (Multiplication Law in Dn ). If two elements of Dn are written in factored form (19), the group operations take the form (a) (k ) (r s ) = (b) (k )-1 =
k+(-1) r
+s
-(-1) k -
for k, r Zn and , s Z2 .
k-r As for uniqueness, the identity k = r s implies = s- . Unless both exponents are congruent to zero, the transformation on the left has determinant +1 while the one on the right has determinant -1, which is impossible. We have now established that the subgroups N = and H = have the properties (i) N H = (e), (ii) N H = Dn , and (iii) N is normal in Dn . Conjugation of an element n N by an arbitrary element in Dn has the following effect.
(20)
gk g -1 = (r s )k (r s )-1 = r ( s k -s )-r = r (
(-1)s k
(-1) )-r =
s
k
| Thus Dn is a semidirect product Zn Z2 under the action H N N in (20). =
6.2.8 Example. In G = Dn the geometric meaning of the elements k and the reflection is clear, but what about products of the form k , the elements in the "other" coset in Dn = N N ? We will show that these are reflections across the lines of reflection symmetry shown in Figure 6.3A: k = (reflection across the line (k
)) 2
where () is the line through the origin making an angle with the +x-axis. Discussion: First consider the element . Its action on basis vectors is shown in Figure 6.3B below. Now compare this with the action of the reflection r(/2) across the line through the origin that makes an angle /2 with the +x-axis. Both have the same action on basis vectors, so = r(/2) .
19
Figure 6.3A. Lines of reflection symmetry of an n-gon centered at the origin. (a) The situation for n odd: every line through a vertex passes through the midpoint of an opposite edge. (b) For even n we get two families of lines, passing through two opposite vertices or through the midpoints of opposite edges.
Figure 6.3B. The action on basis vectors of the operations and reflection r(/2) across a line making an angle /2 with the +x-axis.
The same argument applies to rotation by any angle, in particular the angles k ( = 2/n) corresponding to vertices of the n-gon, therefore k = k = r(k(/2)) for all k Z
This accounts for all lines of reflection symmetry shown in Figure 6.3A, but there is a curious distinction between the cases n = even and n = odd. As in Figure 6.3A, for even n there are two families of lines of reflection: lines (k) through opposite vertices, and lines ((k + 1 )) through midpoints of opposite edges. 2 For odd n a radial line through a vertes also passes through the midpoint of the opposite edge. As we increment k = 0, 1, 2, . . . in this case the lines (k(/2)) cycle twice through the lines of reflection shown in Figure 6.3A(a). In either case, all lines of reflection symmetry are accounted for as elements of the coset N . 6.2.9 Exercise. In Dn show that the group element k -k (with k 0 (mod n)) /
is the reflection L across the line through the origin that passes through the k th vertex of the n-gon. (Label vertices 0, 1, 2, . . . , n - 1 counterclockwise starting with the vertex 20
on the positive x-axis). Hint: Use the results in 6.2.8 *6.2.10 Exercise. Determine the center Z(Dn ) = {x Dn : gxg -1 = x for all g G} for n 3. Note: The answer will depend on whether n is even or odd. When n = 2 the group is abelian and the center is all of D2 . Hint: By 3.1.46 an element x = i j is in the center of Dn x commutes with the two generators and of Dn . This simplifies calculation of the center. 6.2.11 Exercise. Determine all conjugacy classes in D6 and in D7 . Note: Recall that N (hence also its other coset N ) are unions of whole conjugacy classes, so you might start by determining the classes that lie in the normal subgroup. 6.2.12 Exercise. Use Exercise 6.2.11 to determine all normal subgroups in D6 and in D7 . Hint: Recall 5.4.2 and 5.4.6. *6.2.13 Exercise. For n 3 determine the conjugacy classes in Dn . Start by discussing the class C = {gg -1 : g Dn }. Note: The answer is different for odd and even n. D2 Z2 Z2 is abelian and has trivial = classes. 6.2.14 Example (Invertible Affine Mappings Aff(V )). The affine mappings on a finite dimensional vector space V are the maps of the form (21) T (v) = A(v) + a where A is a linear map and a V
It is easily seen that T : V V is an invertible bijection its linear part A is invertible, with det(A) = 0. It is also clear that the set Aff(V ) of invertible affine maps is a group under composition of operators since the composite (22) T T (v) = A (T v) + a = A (Av + a) + a = A A(v) + (A a + a ) = B(v) + b
is again invertible and affine. Any affine map is a composite T = ta A where A is a linear operator on V and ta : V V is the pure translation operator ta (v) = v + a. The components A, a in (21) are uniquly determined, for if A v + a = Av + a for all v then we have A v - Av = a - a . Taking v = 0 we get a = a, and then A v = Av for all v, so A = A as operators on V . Within G = Aff(V ) we find two natural subgroups Translations: N = {ta : a V }. This subgroup is abelian because ta ta = ta+a and since a = a ta = ta , the map j : a ta is an isomorphism between (V, +) and the subgroup N in (Aff(V ), ). Purely Linear Operators: GL(V ) = {A : det(A) = 0}. This is the set of all invertible linear operators A : V V , which is obviously a subgroup in Aff(V ). We have just remarked that G = N GL(V ). It is also clear that N GL(V ) = {I}, where I is the identity operator on V . In fact, every linear operator A leaves the origin in V fixed, but a translation ta does this only when a = 0 and ta = I. We claim that N is a normal subgroup in G, and hence we have a semidirect product | Aff(V ) = N GL(V ). To see that T ta T -1 N for any T Aff(V ) we first apply (21) to compute the inverse of any operator T v = Av + b. (23) T -1 v = A-1 (v - b) = A-1 (v) - A-1 (b)
21
(Simply solve w = T v = Av + b for v in terms of w.) Next we compute the effect of conjugation by T on a translation ta in the special case when T = A is linear and b = 0. We get (24) Ata A-1 = tA(a) for all a V, A GL(V )
because for every v we have Ata A-1 (v) = A(A-1 (v) + a) = AA-1 (v) + A(a) = v + A(a) = tA(a) (v) Thus, conjugating a translation ta by a purely linear operator produces another translation as expected, but it is also useful to observe that When we identify (V, +) N via the bijection j : a ta , the action GL(V ) = N N by conjugation (which sends ta Ata A-1 ) gets identified with the usual action of the linear operator A on vectors in V (which sends a A(a)). Now let T be any element in G. To see that T N T -1 N for all T G, we write T = tb A and compute the effect of conjugation; since t-1 = t-b , equation (24) implies b that T ta T -1 = tb (Ata A-1 )t-b = tb tA(a) t-b = tb+A(a)-b = tA(a) Compare with (24): throwing in the translation component tb has no effect when T acts by conjugation on the translation subgroup N . That makes sense: N is abelian and acts trivially on itself via conjugation.
Figure 6.4. In (a) we show the reflection symmetry L across a line L passing through the midpoint of an edge of the n-gon. (b) The 2-gon degenerates to a line segment with two vertices.
6.2.15 Exercise. Consider the lines of reflection symmetry of the regular hexagon shown in Figure 6.4(a). To whiich group elements in D6 do the reflections r1 and r2 correspond? 6.2.16 Exercise (The Degenerate case n = 2). When n = 2 the n-gon degenerates into a line segment, as shown in Figure 6.4(b). This shape has D2 as its full symmetry group; notice that = . (a) Identify the geometric meaning of each element e, , , in D2 . (b) Verify that D2 is abelian. (c) The only groups of order 4 are Z4 and Z2 Z2 . Which one is D2 ? We close this section with a few case studies. In the first we shall determine all groups of order |G| = 6, up to isomorphism. We will analyze some more complicated groups at 22
the end of Section 6.3 after introducing an important new tool, the Sylow Theorems. 6.2.17 Example (Groups of order |G| = 6). By Cauchy's theorem 4.3.4 there exist cyclic subgroups H2 and H3 of order 2 and 3. Since H2 H3 is a subgroup in both H2 and H3 , by Lagrange it must be trivial with H2 H3 = (e). The subgroup H3 is normal because if gH3 g -1 = H3 for some element g G, then the intersection gH3 g -1 H3 would be trivial (by Lagrange) and the product set would have cardinality |gH3 g -1 H3 | = |gH3 g -1 | |H3 | =9 |H3 gH3 g -1 |
which exceeds the size of G. Contradiction. Once we know H2 Z2 and that H3 Z3 is normal, G must be a semidirect = = product of H Z2 acting on N Z3 . The possible actions are determined by the = = homomorphisms : Z2 Aut(N ). Since N = {e, b, b2} has just three elements, and any automorphism maps e to itself, the only possibilities for an automorphism are (b) = e. Then (b2 ) = e and is the identity map I on N (b) = b2 = b-1 . Then (bk ) = b-k and is the inversion map J(x) = x-1 Thus Aut(N ) = {I, J} with J 2 = I. Write a for the generator of H = {e, a}. A homomorphism : H (Aut(N ), ) is determined by where it sends this generator, and in this example there are just two possibilities. Group G(1) : Take (a) = I. Then (e) = I and (a) = I
so all elements of H go to the identity map on N and the action of H on N is trivial. The multiplication rule for the semidirect product G(1) is (bi , aj ) (bk , a ) = =
(bi (aj )(bk ), aj a )
(bi+k , aj+ )
which is precisely the multiplication law for the direct product. Thus G(1) = N H Z3 Z2 , which by Chinese Remainder is Z6 . = = Group G(2) : Take (a) = J. Then (e) = I and (a) = J (so (ak ) = J k )
and the multiplication law in G(2) is (25) because (bi , aj ) (bk , a ) = (bi , aj ) (bk , a ) = (bi b(-1) k , aj+ )
j
(bi (aj )(bk ) , aj a ) j = (bi J j (bk ), aj+ ) = (bi b(-1) k , aj+ )
j
= (bi+(-1) k , aj+ ) Here (a)j (bk ) = b(-1) k because each successive application of (a) to bk reverses the sign of the exponent.
j
23
The semidirect product in which H Z2 acts by inversion on N Zn is precisely the = = dihedral group Dn , so G(2) D3 . This can also be seen by observing that the elements = = (b, e) satisfy the identities o() = 3 o() = 2 -1 = -1 and = (e, a)
characteristic of the dihedral groups because = = (e, a)(b, e)(e, a) = (eJ(b), ae)(e, a) = (b-1 , a)(e, a) (b-1 J(e), aa) = (b-1 , e) = -1
There are no other possibilities for groups of order 6. In particular, up to isomorphism the only abelian group of order 6 is Z6 . Incidentally, we have indirectly proved that the permutation group S3 is isomorphic to D3 because both are noncommutative and of order 6. One can also write the multiplication law for G(2) D3 in additive form. Under the = isomorphisms 1 : Z3 N 2 : Z2 H 1 ([i]3 ) = bi 2 ([k]2 ) = a
i
for [i]3 Z3 for [k]2 Z2
the inversion map J(bi ) = b-i becomes the additive inversion map J([i]3 ) = -[i]3 on Z3 . Then we can identify G(2) as Z3 Z2 equipped with the following multiplication law.
| 6.2.18 Exercise. Verify that the multiplication law in G(2) = Z3 Z2 is given by
([i], [j]) ([k], []) =
( [i]3 + J j ([k]3 ) , [j + ]2 ) = ( [i + (-1)j k]3 , [j + ]2 )
Note the parallel between this and the multiplicative version (25). The identities in Exercise 6.2.18 are precisely those satisfied by the exponents in the multiplicative law (25). This correspondence is an immediate consequence of the exponent laws. *6.2.19 Exercise. Can you devise a bijective map : S3 D3 that effects the isomorphism mentioned above? Hint: What subgroups in S3 might play the roles of N = {e, , 2 } and H = {e, } in D3 ?
| In any discussion of general semidirect products N H we have to address the following tasks
Determine the elements in Aut(N ). This can be difficult, but we already know how to do it if N Zn is cyclic because we have shown that Aut(Zn , +) (Un , ), the = = multiplicative group of units in the ring (Zn , +, ) (see 3.5.3). Determine all homomorphisms : H Aut(N ). Here we use the fact that a homomorphism is completely determined once we know where it sends the generators of H. This is particularly simple if H = a is a cyclic group, with just one generator. However, not all assignments sending a to an automorphism a Aut(N ) yield valid homomorphisms when we assign (26) (ak ) = ((a)) = k a 24
k
(k Z)
If o(a) = n, so an = e, then (a )n = I (the identity map on N ) so the map (26) is a homomorphism only when the assigned automorphism satisfies this "consistency condition." In turn, (a )n = I means that the order o(a ) as an element in Aut(N ) is a divisor of n = o(a).
| Define a semidirect product group G() = N H.
(27)
(n, h) (n , h ) = (nh (n ) , hh )
for all (n, h) N H
In a previous example we ended up having to determine all possible homomorphisms : (Z2 , +) Aut(Z3 , +) (U3 , ) = That was easy because the groups were quite small, and both were cyclic. In what follows we provide an explicit procedure for determining all semidirect products in the | (cyclic)(cyclic) case. Life gets more interesting if H has several generators, but there are some general guidelines for determining the assignments : (generators) Aut(N ) that yield homomorphisms : H Aut(N ). We have already mentioned one constraint on such assignments: If a H has order o(a) = n and (a) = a in Aut(N ), the order o(a ) of the automorphism must be a divisor of n = |H|. Other useful constraints on homomorphisms : H Aut(N ) are forced by the Lagrange theorem. The order of the image group (H) must divide the order of Aut(G). The order of the image group (H) must also divide the order of H. The first follows because (H) is a subgroup in Aut(N ); the second follows from Lagrange and the First Isomorphism Theorem because (H) H/ker() and = |H| = |ker | |H/ker | = |ker | |(H)| Sometimes these conditions by themselves force to be trivial, with (h) = idN for all h H, and then the direct product N H is the only possibility.
| The Cyclic Case Zn Zm . Since we are writing N and H in additive form it will be convenient to take the obvious additive generators for N = Zn and H = Zm , namely b = [1]n and a = [1]m . When N = (Zn , +) we know its automorphisms. As shown in 3.5.3,
Aut(Zn , +) is isomorphic to the group (Un , ) of multiplicative units in Zn This isomorphism is effected by the map : Un Aut(Zn ) that takes a multiplicative unit [r]n to the "multiplication operator" [r]n : [i]n [r]n [i]n = [ri]n for [i]n Zn
Therefore, finding a homomorphism : Zm Aut(Zn ) is equivalent to finding a homo~ morphism : (Zm , +) (Un , ). We work through the details below.
| Notation in Semidirect Products Zn Zm . Describing the group law in terms of the Cartesian product space Zn Zm can be a bit confusing since integers are combined
25
using the (+) operation within Zm and Zn ; multiplied via () within the group of units Un ; and also multiplied in forming the actions [r] : [] [r][] = [r] for [r] Un , [] Zn
6.2.20 Proposition. If : (Zm , +) (Un , ) is the homomorphism that sends the generator a = [1]m of Zm to an element b = [r]n in Un that satisfies the compatibility condition bm = [1]n , then the corresponding group operation in the semidirect product | Zn Zm takes the form (28) ([i]n , [j]m ) ([k]n , []m ) = ([i + rj k]n , [j + ]m )
Proof: Since Zm is an additive group the j th "power" of the generator a is j a = a + . . . + a, and since is a homomorphism from an additive group to a multiplicative group (Aut(N ), ) we get ([j]m ) = (j a) = = = (a + . . . + a) (a) . . . (a)
j (a)j = [r] = [r]j = [rj ]
Hence, (suppressing some subscripts m, n for clarity), we get ([i]n , [j]m ) ([k]n , []m ) = = = as required We now turn to various other computed examples. 6.2.21 Example. Determine the group of units (U5 , ) and its isomorphism type. Then find all possible homomorphisms : Z3 U5 . Describe the possible semidirect products | G = Z5 Z3 . Discussion: Obviously Aut(Z5 , +) U5 = {[k] : gcd(k, 5) = 1} = {[1], [2], [3], [4]} = and |U5 | = 4. There are two possibilities: U5 Z4 and U5 Z2 Z2 . If we calculate = = the orders of a few elements in U5 we find that o([2]) = 4, so (U5 , ) (Z4 , +). = If : Z3 U5 is a homomorphism, the size |(Z3 )| of the image subgroup in U5 must be a divisor of both 3 and 4. Since gcd(3, 4) = 1, the range of must be the trivial subgroup {[1]} in U5 , so the only action Z3 Z5 Z5 by automorphisms is the trivial action. The direct product Z5 Z3 Z15 is the only possible semidirect product. = (Notice that we didn't need to know the structure of U5 to see this!) 6.2.22 Example. Repeat the last example taking N = Z9 and H = Z3 . Discussion: Now Aut(N ) = U9 = {[1], [2], [4], [5], [7], [8]} and |U9 | = 6. Up to isomorphism there are just two groups of order |G| = 6, and since the group of units is abelian we see that U9 (Z6 , +), without inspecting the orders of any elements in U9 . [ In fact, = o([2]) = 6 so a = [2] is a cyclic generator of U9 .] However, it is useful to actually write down the orders o(x) of elements in U9 and the groups x they generate, see the Table below. 26
([i] + (a)j ([k]), [j] + []) j = ([i] + [r] ([k]) , [j] + [])
([i] + [r]j [k], [j + ]) ([i + rj k]n , [j + ]m )
x Table. Orders of elements in U9 and the cyclic groups they generate. [1] [2] [4] [5] [7] [8]
o(x) 1 6 3 6 3 2
x 1 1, 2, 4, 8, 16 7, 14 5 1, 4, 16 7 1, 5, 25 7, 35 8, 40 4, 20 2 1, 7, 49 4 1, 8 -1
With this we can determine the possible homomorphisms : Z3 U9 . Our only flexibility in defining lies in specifying where sends the generator a = [1] of H = Z3 . Since a has order 3, can only send a to an element b U9 such that b3 = e, and there are just three such elements: b = [1], [4], [7]. Each assignment yields an action and a semidirect product structure on Z9 Z3 Group G(1) : sends a to [1] U9 . Then ([j]3 ) = (j a) = [1]j = [1] for all j. The action of Z3 on Z9 is trivial and we get the direct product G(1) Z9 Z3 (which, incidentally, is not Z27 . Why?). = = Group G(2) : sends a to the element [4] in U9 , which corresponds to the automorphism [4] : [k] [4][k] = [4k] for all [j] Z9 . By (28), the group law in the semidirect product is ([i]9 , [j]3 ) ([k]9 , []3 ) = ([i + 4j k]9 , [j + ]3 ) Group G(3) : sends a to the element [7] in U9 , which corresponds to the automorphism [7] : [k] [7][k] = [7k] for all [j] Z9 . By (28), the group law in the semidirect product is ([i]9 , [j]3 ) ([k]9 , []3 ) = ([i + 7j k]9 , [j + ]3 )
| The preceeding analysis shows that all semidirect products Z9 Z3 are included in our list of groups G(1) , . . . , G(3) ; it does not, however, insure that the groups we have constructed are distinct up to isomorphism. Since [7] = [4]-1 in (U9 , ) there seems to be a strong symmetry between the multiplication laws in G(2) and G(3) . In fact, these groups are | isomorphic, so there is only one nonabelian semidirect product of the form Z9 Z3
*6.2.23 Exercise. Verify that the map : Z9 Z3 Z9 Z3 given by ([i], [j]) = ([1], [2j]) is an isomorphism from G(3) to G(2) . Note: is obviously a bijection on the Cartesian product space Z9 Z3 because [2] is a unit in Z3 . You might wonder how anyone ever came up with the map that effects this isomorphism. Here is a clue: The products in G(2) and G(3) involve multipliers of the form 4j and 7j . The equation 4x = 7 in U9 has x = 2 as a solution, and therefore 42j 7j in Z9 for all j Z3 .
| *6.2.24 Exercise. Describe all semidirect products Z3 Z3 by finding all possible homomoprhisms : Z3 (U3 , ) Aut(Z3 , +). =
*6.2.25 Exercise. If p, q > 1 are primes and p q, prove that the only semidirect 27
| products Zq Zp are trivial direct products. Produce an example showing that this is not necessarily true if p < q.
The next example involves an N that is abelian but not cyclic. 6.2.26 Exercise. Describe all automorphisms of the non-cyclic group N = Z2 Z2 . Then describe all homomorphisms : Z2 Aut(N ), and determine all possible semidi| rect products N Z2 . S3 , the group of permutations of 3 objects. What 3 objects could possiHint: Aut(N ) = bly be permuted by a typical automorphism of Z2 Z2 ? Which elements S3 satisfy 2 = e? What are the corresponding automorphisms of Z2 Z2 ? 6.2.27 Exercise. Make a table showing o(x) for all x U16 . (a) Use this to prove that U16 Z4 Z2 . = (b) Explain why there are no isomorphisms between the groups Z8 , Z4 Z2 , and Z2 Z2 Z2 . | (c) Determine all semidirect products Z16 Z7 .
| (d) Determine all semidirect products Z16 Z2 .
The next two exercises on product groups will play an important role in a later discussion (Section 6.3) in which we shall describe all groups of order |G| = 8, a tricky case whose outcome is quite different from that when |G| = 6. 6.2.28 Exercise. If G is a group such that o(x) = 2 for all x = e, so x2 = e for all x, prove that G is abelian. Hint: Recall the discussion of 6.1.17 where we proved that all groups of order |G| = 4 are abelian. 6.2.29 Exercise. If G is a finite group such that o(x) = 2 for all x = e, (a) Prove that |G| = 2n for some n N. (b) Use 6.2.28 to prove that G Z2 . . . Z2 = (n factors)
Group Extensions. Suppose N G is a normal subgroup. Then there is a natural exact sequence of homomorphisms (29) e - N - - - G - - - --- - - - H = G/N - e
1 =id 2 =
where : G G/N is the quotient map. Here exact means range(i-1 ) = ker(i ) at every step in the sequence, which is certainly true in (29) since idN is one-to-one, is surjective, and ker() = N . The middle group G in such a sequence (29) is called an extension of the group G/N by the group N . In some sense G is a composite of N and H = G/N , but additional information is needed to know how they are put together. Part of this information is obtained by noting that there is a natural group action GN N , given by g n = g (n) = gng -1 . This makes sense because N G. For each g G the operator g : N N is actually an automorphism of N , and it is easily checked that (30) The map : g g is a homomorphism from G to the group of automorphisms Aut(N ), so that e = idN and g1 g2 = g1 g2 .
For any g the conjugation operators g (x) = gxg -1 are inner automorphism of G (recall Section 3.5). However, the restrictions g = g |N are not necessarily inner automorphisms of N because there might not be any element b N such that conjugation by 28
b matches the restricted action of g on N . For instance, if N is abelian all inner automorphisms are trivial; but the restrictions g = g |N can be nontrivial because g lies outside of N . Sometimes, if we are lucky, the sequence (29) splits: there is a subgroup H G that cross-sections the G/N cosets in the following sense. Then the action (30) is all we need to solve the problem of reassembling G from its components N and G/N : the semidirect product construction does the job. 6.2.30 Definition. Let G be a group and N a normal subgroup, as in (29). A subgroup H is a cross-section for G/N if each coset in G/N meets the set H in a single point; in particular, H N = (e). If such a subgroup exists we say that the sequence (29) splits. Such subgroups are generally not unique. The meaning of the cross-section property is shown in Figure 6.5. The idea is that Every coset C in G/N has a unique representative x (i.e. xN = C) such that x lies in the cross-section subgroup H. Existence of such a representative follows because the coset C meets the set H; uniqueness follows because there is just one point in C H. What all this means is that the quotient map : G G/N restricts to H to give a bijective homomorphism H : H G/N , which implies that H G/N for any cross-section. In effect, = cross-sections H are copies of the quotient group G/N embedded back inside G. Various conditions are equivalent to existence of a cross-section. We summarize the possibilities in the next lemma.
Figure 6.5. A cross-section H meets each coset in G/N in a single point, so the quotient map restricts to an isomorphism from H to G/N , as shown.
6.2.31 Lemma. Let G be a group, H a subgroup and N a normal subgroup. Then the following statements are equivalent (a) The product set N H is equal to G and N H = (e) (b) Each g G has a unique factorization g = nh with n N, h H. (c) H is a cross-section for G/N cosets These conditions are satisfied precisely when the sequence (29) splits, and then we have H G/N . = Remark: Notice that N H = HN by normality of N , because every product nh can be rewritten as nh = h(h-1 nh) = h n . Thus each g also has a unique factorization as g = h n . To avoid notational mess later on we restrict attention to factorizations g = nh in which the normal element lies on the left, even though cosets in G/N have the form gN . Proof: For (c) (a), if H cross-sections cosets we get H N = (e) by looking at the coset eN = N . For any x G the intersection xN H = (b) consists of a single point such that xN = bN . In particular there is some n N such that x = bn HN = N H. Hence G = N H. For (a) (b), every group element has some decomposition g = nh because G = N H. If g = nh = n h , then n-1 n = h(h )-1 lies in N H = (e), so that n = n and h = h . The decomposition is unique. For (b) (c), unique factorization implies that g = nh = h(h-1 nh) hN . Hence gN = hN , so every coset in G/N has at least one representative in H. If we could find 29
two representatives h, h H gN , we would then have hN = gN = h N h = hn for some n N h-1 h = n H N = (e) n = e and h = h Therefore each coset meets H in a single point, as required.
| Obviously condition (a) means G is a semidirect product of the form N H, and we can reconstruct G once we know how elements h H act as automorphisms h = h |N on the normal subgroup N . In the next section we will see that among groups of order |G| = 8 there is one (the group Q8 of unit quaternions ) that is a non-split extension of G/N Z2 by N Z4 . For non-split extensions an entirely new theory of group = = cohomology is needed to see how the subgroups N and G/N combine to reconstruct G. Here is an example of a non-split extension. Although it involves an infinite group, it illustrates the sort of obstructions that can prevent the existence of a subgroup that cross-sections the N -cosets in G. The quaternion group Q8 to be discussed in Section 6.3 is a finite group exhibiting similar behavior.
6.2.32 Example. If we take G = R and N = Z, the extension e Z R G/N e does not split, R - (S 1 , ) so G is not a semidirect product of N and H. To see why, recall that G/N is isomorphic to the circle group ~ S 1 = {z C : |z| = 1}. In fact (as in 3.3.3), the homoR/Z 1 morphism : (R, +) (S , ) given by the exponential ~ map (t) = e2it has kernel ker = Z, and hence by the Figure 6.6. The induced map is ~ an isomorphism such that = First Isomorphism Theorem 3.1.13 the map factors (diagram commutes). through the quotient map : R R/Z to give an iso~ morphism : (R/Z, +) (S 1 , ) as shown in Figure 6.6 at right. Arguing by contradiction, we now show that no subgroup H in R can cross-section the cosets x + Z in R/Z. Suppose such an H actually exists. Consider any rational value 0 < < 1. Then m Z for some m N, and hence 1 = (m) = ()m . Since H is a cross-section for R/Z-cosets, there is some x H such that x + Z = + Z. Therefore x - Z, mx - m Z, and then (mx) = (m) = 1 because ker = Z. But if H is a cross-section we also know that the restricted homomorphism H : H S 1 is a bijection, and hence an isomorphism of groups. Since x H mx = x + . . . + x H, the only way to get (mx) = 1 is to have mx = 0 because : H S 1 is one-to-one and we already have (0) = 1. This in turn implies that x = 0, which is impossible because it would imply that x 0 (mod 1) and hence that is an integer contrary to our choice of 0 < < 1. Conclusion: no subgroup can cross-section the cosets in R/Z.
30
6.3 The Sylow theorems.
A group is called a p-group if its order is some power ps of a prime p > 1. If G is a nontrivial finite group, its order |G| = n will have various prime divisors pi > 1 r with multiplicities ni 1 such that |G| = i=1 pni . We now show that the pattern of i subgroups in G is keyed to these prime divisors and their multiplicities. The connection is revealed in the three "Sylow Theorems" presented below. 6.3.1 Definition. Let G be a finite group whose order has n = i=1 pni as its prime i factorization. A subgroup H G is a pi -group if its order is some power of pi . It is a Sylow pi -subgroup if its order is as large as possible, namely |H| = pni . For any i prime p > 1 we write Sylp (G) to indicate the collection of all Sylow p-subgroups in G. Unless p is a divisor of |G| this collection will be empty; if Sylp (G) = , it might contain several distinct Sylow p-subgroups. The main point of the Sylow theorems is that Sylp (G) is not empty if p divides |G|, and this observation is the starting point if we wish to unravel the structure of a given finite group. 6.3.2 Theorem (The Sylow Theorems). Let G be a group of finite order n = r ni i=1 pi . Then for each prime factor pi we have (a) G contains a subgroup Spi which has exactly pni elements. i (b) If Spi is a fixed Sylow pi -subgroup, any subgroup H whose order is a power of pi can be conjugated to lie within Spi i.e. there is some g G such that gHg -1 Sp . In particular all Sylow pi -subgroups are conjugates of one another. (c) The number of distinct Sylow pi -subgroups is equal to 1 + mpi for some m, so their number is congruent to 1 (mod pi ). The number of pi -Sylow subgroups must also be a divisor of |G|. For each prime divisor p > 1, all the Sylow p-subgroups are isomorphic since they are conjugates of each other. Proof: For (a) we start with the abelian case, working by induction on n = |G|. If n = 1 the theorem is vacuous, true by default. For n = 2 we have G Z2 ; p = 2 is the only = divisor and G itself is the Sylow 2-subgoup. So, we may assume n 2 and p > 1 is one of its prime divisors, say with pk the largest power dividing n. In Cauchy's theorem 4.3.5 we showed that G contains a cyclic subgroup H = a of order |H| = p. Since G is abelian, H is normal; the abelian quotient G = G/H has order less than n, so we may apply the induction hypothesis to it. By Lagrange's theorem 3.4.1, if p is not a divisor of |G/H| then p has multiplicity k = 1 and H is the Sylow p-subgroup we seek. Otherwise, by Lagrange's theorem we have |G| = |G/H| |H| = p|G/H|, so p is a divisor of |G/H| with multiplicity k - 1 1. By the induction hypothesis G/H contains a Sylow p subgroup H, of order pk-1 . The pullback H = -1 (H) = {g G : (g) H} under the quotient homomorphism : G G/H must have order |H | = pk since |H | = |H| |H|. Thus H is a Sylow p-subgroup for G, which proves the abelian version of (a). For general groups G we again use induction on n = |G| to prove (a), and again the cases n = 1, 2 are trivial (not to mention abelian). So, assume n 2 and that p > 1 is a prime divisor whose largest power in n is pk . If G is nonabelian, the decomposition of G into conjugacy classes is nontrivial. The class equation 4.3.1 says |G| = |Z(G)| +
xS r
|Cx | = |Z(G)| +
xS
|G| |ZG (x)|
where S is a set of representatives for the nontrivial conjugacy classes Cx in G, and 31
ZG (x) = {g G : gx = xg} is the centralizer of x (the stabilizer StabG (x) when we let G act on itself by conjugation). If there is a nontrivial class Cx such that |ZG (x)| is divisible by pk , then pk is also the largest power of p that can possibly divide |ZG (x)| since ZG (x) is a subgroup in G. Clearly |ZG (x)| < |G| because Cx is nontrivial, so by the inductive hypothesis this subgroup already contains a Sylow p-group for G. If no such x exists, then for every nontrivial class we have #(elements in the class) = |G| |ZG (x)| is divisible by p .
because the multiplicity of p in the centralizer is less than that in G. Therefore by the class equation, the number of elements in the center Z(G) must also be divisible by p and hence |Z(G)| must include a factor of the form ps for some 1 s k. If s = k we simply apply the abelian result to Z(G) and are done. Otherwise, consider the quotient group G/Z(G)p where Z(G)p is the Sylow p-subgroup of the center, with |Z(G)p | = ps . (Any subgroup of the center Z(G) is normal in G, so the quotient is a group.) Induction applies to this quotient, whose order involves the prime factor pk-s . Therefore the quotient contains a Sylow p-subgroup S p of this order. The pullback Sp in G has order pk since we factor out a group of order ps to get S pi . This Sp is the desired Sylow p-subgroup in G. To prove (b) and (c) we use the following instructive lemma about actions of p-groups. 6.3.3 Lemma. Let X be a finite set acted on by p-group G. Let X G = {x X : g x = x, all g G} be the set of G-fixed points in X. Then |X G | |X| (mod p). In particular fixed points must exist if |X| 0 (mod p). Proof: The set X G is obviously G-invariant and so is the difference set X X G , which must then be a union of disjoint nontrivial G-orbits. The cardinality |G|/|StabG (x)| of any such orbit is a divisor of |G| and so must be a power ps with s 1. Hence |O| is congruent to 0 (mod p) for every nontrivial orbit O in X. But then the same must be true for the union X X G of these orbits. That means |X| = |X X G | + |X G | |X G | as claimed. For (b) if |G| = pm the group is already a Sylow subgroup and there is nothing to prove. Otherwise, let n = pk m with gcd(m, pk ) = 1. Fix a Sylow p-group Sp and let H be a subgroup whose order is a power of p. Consider the permutation action of H on the coset space X = G/Sp . Then |Sp | = pk and |G/Sp | = m 0 (mod p), so by 6.3.3 there is at / least one fixed point xSp in X. But then HxSp = xSp x-1 HxSp = Sp x-1 Hx Sp and we're done. For (c) we fix a Sylow p-subgroup Sp and look at the group action Sp X X of Sp by conjugation on the set X = Sylp of all Sylow p-groups in G. The base point Sp is a fixed point in X; we claim that there are no others, and then (c) follows from our lemma. If there were another fixed point H X it would be normalized by Sp in the sense that gHg -1 = H for g Sp . Thus Sp is contained in the normalizer of H: the subgroup N = NG (H) = {g G : gHg -1 = H}. Hence Sp and H are Sylow p-subgroups of N , and by (b) there exists some n N such that Sp = nHn-1 . By definition of N this makes Sp = H as claimed. Let P be any base point in Sylp (G). By (b) the action G Sylp Sylp is transitive so |Sylp | = |G|/|StabG (P )| and |G| is a multiple of |Sylp |, as claimed. In 4.3.5 (Cauchy theorem) we proved that if p is a divisor of n = |G| then there are 32 (mod p)
Figure 6.7. If prime divisor p of |G| has multiplicity 1, the Sylow p-subgroups (1) (N) Sp , . . . , S p are "essentially disjoint" as indicated in (a), so their union has cardinality |Ep | = N (p-1)+1. If p, q are different prime divisors, the unions Ep , Eq of the corresponding Sylow subgroups are essentially disjoint, as in (b), even if p and q have multiplicities (i) greater than 1 (in which case the individual Sp might have nontrivial overlap). Thus |Ep Eq | = |Ep | + |Eq | - 1.
elements in G such that o(x) = p. If pk is the largest power dividing n, one can generalize the Sylow theorems to prove that there exist subgroups of order pr for every 1 r k. Note: In analyzing the structure of groups, special interest attaches to the p-groups, for which |G| = pk , so it is worth noting that A finite group G is a p-group the order of each element is a power of p. Implication () is trivial (Lagrange). For the converse (), if some prime q = p appeared in the prime decomposition of |G|, then by 4.3.5 there would be a cyclic subgroup of order q, which is impossible. By 6.3.2(b), a Sylow p-subgroup is normal in G if and only if there is just one such subgroup. Sometimes we can determine when this happens by using 6.3.2(c), and in any case a lot can be learned about the pattern of Sylow subgroups by looking at intersections of conjugates Sp gSp g -1 with Lagrange and the Sylow theorems in mind. For instance, suppose p has multiplicity 1 in the prime factorization of n = |G|. Then the distinct Sylow (1) (N ) p-subgroups Sp , . . . , Sp all have order p and by Lagrange their pairwise intersections are trivial. Hence their union has cardinality N (p - 1) + 1, which cannot exceed |G|. (i) This and the requirement that N 1 (mod p) can put serious constraints on the Sp . The idea is shown in Figure 6.7(a). (i) N Another useful counting principle concerns the unions Ep = i=1 Sp and Eq = (j) M of all the p-Sylow and q-Sylow subgroups in G, where p, q are distinct prime j=1 Sq divisors of |G|. The subgroups in Ep , Eq have orders |Sp | = pk and |Sq | = q where (j) (i) k, N. Hence gcd(pk , q ) = 1, which implies that Sp Sq = (e) for all i, j. That forces the "blobs" Ep and Eq to be essentially disjoint in the sense that Ep Eq = (e), because any x Ep Eq would lie in the intersection of a p-Sylow and a q-Sylow subgroup. Obviously the union of these blobs must fit inside G, so that |Ep Eq | = |Ep | + |Eq | - 1 |G|, see Figure 6.7(b). (There must also be room outside Ep Eq for Sylow subgroups associated with other prime divisors of n.) These constraints also provide useful information about the pattern of Sylow subgroups in G. 6.3.4 Example. Let G be an arbitrary group of order |G| = 28 = 7 22 . If H7 is a 33
(i) (j)
7-Sylow subgroup then |H7 | = 7 and N = #(7-Sylow subgroups) can only equal 1, 8, 15, 22 ( |G| = 28). In fact N = 1 because the next possible value N = 8 would make the union E7 of the Sylow 7-subgroups have cardinality 8(7 - 1) + 1 = 49, which is bigger than the group itself. (Besides, N = 8 can't work because it is not a divisor of |G| = 28.) Thus the 7-Sylow subgroup H7 (Z7 , +) is normal in G. = The 2-Sylow subgroups H2 all have order |H2 | = 22 = 4. As shown in Section 6.2, all groups of order 4 are abelian and up to isomorphism the only possibilities are Z4 (cyclic), and Z2 Z2 . The number of 2-Sylow subgroups can only be N = 1, 3, 5, 7, ...; only 1 and 7 are divisors of |G| = 28, so #(2-Sylow subgroups) is either 1 or 7. Fix a 2-Sylow subgroup H2 . Then H2 H7 = (e) since gcd(4, 7) = 1; it follows that |H7 H2 | = 28 and | | G = H7 H2 . Thus G is a semidirect product H7 H2 = Z7 H2 . Case 1: H2 = Z4 . Identifying H2 = (Z4 , +), consider the standard generator a = [1]4 of Z4 . We must determine all homomorphisms : Z4 (U7 , ) x o(x) Subgroup x = Aut(Z7 , +). Since U7 = {[1], [2], [3], . . . , [6]} is abelian [1] 1 1 and |U7 | = 6 it follows from 6.2.17 that U7 (Z6 , +). = We won't need this specific information below, but what [2] 3 1, 2, 4 we do need is a list of the orders of the elements in U7 , and the groups x they generate, so we can decide which [3] 6 all elements b U7 may be assigned as images (a) of the [4] 3 1, 4, 2 generator of H2 . The orders are listed in the table at 4 right. Since 4 a = [0] in Z4 (a) = [1] in U7 , the [5] 6 all only possible assignments are (a) = [1] or [6] = [-1], [6] 2 1, 6 -1 which correspond to the automorphisms [1] (identity map) and [-1] (inversion) on Z7 .
7 Case 1A: (a) = [1] = idZ7 . This yields the trivial action of H2 = Z4 on N = Z7 ; the corresponding group is the direct product G(1) = Z7 Z4 Z28 . = Case 1B: (a) = [-1] = J (the inversion automorphism on Z7 ). The automorphisms corresponding to the various elements in Z4 = {ja : 0 j < 4} are (ja) = (a)j = J j , so that ([0]) = I ([1]) = J ([2]) = J 2 = I ([3]) = J 3 = J
Data Table for U .
In Proposition 6.2.20 we showed that the multiplication law in the resulting semidirect | | product G(2) = H7 H2 Z7 Z4 takes the form = (31) ([i]7 , [j]4 ) ([k]7 , []4 ) = ([i + (-1)j k]7 , [j + ]4 )
6.3.5 Exercise. Suppose we specify generators a, u and write H2 and H7 in multiplicative notation, so that H2 = {e, a, a2 , a3 } and H7 = {e, u, u2, . . . , u6 }. Prove that the | multiplication law in G(2) = H7 H2 takes the form (ui , aj ) (uk , a ) (32) = (ui (a)j (uk ), aj+ ) = (ui+(-1) k , aj+ )
j
in which the exponents i, k Z7 and j, Z4 satisfy (31). Note: In multiplicative notation, (a) = a is the operator that maps uj to u-j for all 0 j < 7. Furthermore, (ai ) = (a)i . Case 2: H2 = Z2 Z2 . The analysis is complicated by the fact that the acting group is not cyclic. It will be convenient to write H2 multiplicatively, as H2 = {e, u, v, w} where u2 = v 2 = w 2 = e and uv = w, vw = u, wu = v 34
Obviously H2 is the internal direct product of the two subgroups u v Z2 . = = Consequently each element g H2 has a unique factorization in the form g = ui v j with i, j Z2 . As always, a homomorphism : H2 (U7 , ) is determined by where it sends the generators u, v. Furthermore K = ker() can only have cardinality |K| = 1, 2, 4. Case 2A: |K| = 4 or 1. In the first case H2 = K acts trivially on H7 and we have a direct product G(3) = Z7 Z2 Z2 Z14 Z2 . The second case |K| = 1 cannot arise = because would then be injective and (H2 ) would be a subgroup of order 4 in a group U7 of order 6. Case 2B: |K| = 2. Any such must "kill" e and exactly one other element (sending both to [1] in U7 ). By relabeling points in H2 we may assume K = {e, u}; then since w = uv we get (w) = (u)(v) = (v) with (v) = [1] in U7
Since v 2 = e the image (v) can only be an element x in U7 such that x2 = [1], and since (v) = [1] the only choice is (v) = [6] = [-1]. This corresponds to the inversion automorphism [-1] = J on Z7 , so in this case is fully determined: (e) = (u) = I (v) = (w) = J
Let us label the resulting semidirect product as G(4) . Writing both H2 = {ui v j : i, j Z2 } (unique factorization) and H7 = a in multi| plicative form, the multiplication law in G(4) = H7 H2 takes the form (ak , ui v j ) (a , ur v s ) = (33) = = = (ak (ui v j )(a ), ui+r v j+s ) (ak (u)i (v)j (a ), ui+r v j+s ) (ak (v)j (a ), ui+r v j+s ) (since (u) = I) (ak+(-1) , ui+r v j+s )
j
It would seem that we have produced four distinct groups such that |G| = 28, but appearances sometimes deceive. How do we know there isn't some "accidental" isomorphism G(i) G(j) despite the differences in the way the groups are constructed? The = answer is: Without further investigation, we don't! Obviously Z28 Z14 Z2 (why?), = and an abelian group can't be isomorphic to a nonabelian group, so the only interesting possibility is that G(2) G(4) . To disprove this we might compare various structural = properties of the two groups e.g. highest orders of elements, the sizes and isomorphism types of the centers, the number and size of the conjugacy classes, etc all of which can be computed once the group law on the Cartesian product set H7 H2 has been determined. There also seems to be a "missing" group of order 28, namely the dihedral group D14 . Where does it appear in the list G(1) , . . . , G(4) ? A look at the way G(4) was constructed shows that the element u H2 G(4) acts trivially on every element of the normal subgroup H7 . This action is just conjugation by u, u (n) = unu-1 restricted to elements n H7 , which means that u commutes with all n H7 . Since u also commutes with everyone in the abelian subgroup H2 it follows that u commutes with all g G(4) = H7 H2 . Thus the cyclic subgroup U = u Z2 is in = the center of G(4) . Obviously U H7 = (e), so the product set M = U H7 is a normal subgroup of order 14 in G(4) , and is isomorphic to Z2 Z7 Z14 . On the other hand, = every element of G(4) has a unique factorization g = xy with x H7 , y H2 . From this it follows easily that the element v H2 lies outside of M , and since o(v) = 2 the subgroup V = v Z2 has the properties V M = (e), M V = G(4) , M G(4) . This = | | allows us to recognize G(4) as a semidirect product M V Z14 Z2 . = 35
| In fact, G(4) is the missing dihedral group. The possible semidirect products Z14 Z2 correspond to homomorphisms : Z2 U14 Aut(Z14 , +). But U14 = {[1], [3], [5], [9], [11], [13]} = is abelian and has order 6, hence U14 Z6 ; only the elements x = [1] and x = [13] = [-1] = in U14 satisfy the compatibility condition x2 = [1], so the only possible assignments of the generator v V Z2 to elements in U14 are =
(v) (v)
= [1] = idM = [-1] = J (inversion automorphism on M = Z14 )
The first would make G(4) an abelian direct product Z14 Z2 , which is impossible; the other choice clearly yields G(4) D14 . = Is G(2) G(4) ? We have just seen that |Z(G(4) )| 2. On the other hand an element = g = uk a is in the center of G(2) e = xgx-1 for all x = ui aj G(2) . Applying the multiplicative form of the group law in G(2) worked out in equation (32), we get e = = = = (ui aj )(uk a )(a-j u-i ) ui (aj uk a-j )(a u-i a- )a ui a(-1) k (a-(-1) i )a u(-1)
j j
k+[1-(-1) ]i
a
for all i, j. By unique decomposition we must have a = e; hence 0 (mod 4) and [1 - (-1) ] = 0. We then get u(-1)
j
k
=e
and hence
k 0 (mod 7)
Therefore g = uk a = e is the only central element in G(2) and G(2) cannot be isomorphic to G(4) . Abelian groups and the Sylow theorems. If G is a finite abelian group the Sylow theorems provide an explicit natural direct product decomposition. This is not the final answer if we want to know the detailed structure of finite abelian groups, but it is a big step in that direction. The proof uses the observation that in an abelian group all Sylow p-subgroups are automatically normal subgroups, and hence by Theorem 6.3.1(b) there is just one Sylow p-subgroup for each prime divisor of the order |G|. 6.3.6 Theorem. Any finite abelian group is isomorphic to a direct product Sp1 . . . Spr where n = r pni is the prime decomposition of the order |G| = n and Spi i=1 i is the unique Sylow pi -subgroup in G of order pni . This direct product decomposition i is canonical: the subgroups Spi are uniquely determined, as are the primes pi and their exponents ni . Note: The components Spi need not be cyclic groups. For instance, if p = 2 we might have S2 = Z2 Z4 which cannot be isomorphic to the cyclic group Z8 of the same size. Determining the fine structure of the components Spi would require further effort, leading to the Fundamental Structure Theorem for finitely generated abelian groups. The coarse decomposition 6.3.6 is the first step in that direction. Proof: Let's write Si for the unique Sylow pi -subgroup. For each index 1 i r the i product set Hi = j=1 Sj is a normal subgroup (G is abelian). We now apply Lagrange's n theorem and the counting principle 3.4.7 to show that its order is |Hi | = i pj j . j=1 n1 Obviously |H1 | = |S1 | = p1 . When i = 2, the order of the subgroup H1 S2 = S1 S2 must divide both pn1 and pn2 , and hence the intersection H1 S2 is trivial; it follows 1 2 immediately from 3.4.7 that H2 = H1 S1 has cardinality |H1 | |S2 |/|H1 S2 | = pn1 pn2 1 2 36
At the next stage we have H3 = H2 S3 and H2 S3 is again trivial because these subgroups have different prime divisors; applying 3.4.7 we get |H3 | = |H2 | |S3 | = pn1 pn2 pn3 . 1 2 3 Continuing inductively we prove our claim. This already implies that G is a direct product. In fact, since |G| = |Hr | we see that G is equal to the product set S1 S2 . . . Sr . It remains only to check that if a1 a2 . . . ar = e with ai Si , then each ai = e. Let q be the smallest index such that a non-trivial decomposition of the identity occurs. Certainly q > 1 and aq = e, and then a-1 = q a1 . . . aq-1 . But on the left we have an element of Sq and on the right an element of the subgroup Hq-1 . These subgroups have trivial intersection, which is impossible if aq = e. Thus every element in G has a unique decomposition of the form a1 a2 . . . ar , and G is the direct product of its Sylow subgroups. Essentially, this result says that to understand the the internal structure of any finite abelian group it suffices to analyze the abelian p-groups those with just one prime divisor and order pk for some k In order for the Sylow subgroups to be useful indicators of the overall structure for noncommutative G it is necessary to prove that they are pervasive in G. Here's what that means. 6.3.7 Lemma. Let G be a nontrivial finite group and let S be the subgroup generated by all the Sylow subgroups in G: S= {H : H Sylpi (G), 1 i r }
where the pi > 1 are the prime divisors of |G|. Then S is all of G. Proof: The result has already been established when G is abelian. We argue by induction on n = |G|. The result is trivial when n = 2, so we may assume that n > 2 and that the theorem is true for all groups of order at most n - 1. If S = G, there exist elements a S. The cyclic subgroup M = a must have order / |M | = m that divides |G| = n, which means that only the pi can appear in the prime factorization of m. Let p be one of those primes. Any Sylow p-subgroup for M will have order ps for some s ni (if p = pi ). By Theorem 6.3.1(b), every such subgroup must be contained in one of the Sylow p-subgroups for G. The product of the Sylow subgroups in M is therefore contained in S. But by its definition M is abelian and hence (by 6.3.5) is the product of its Sylow subgroups, which means we have a M S. That is impossible since we are assuming a lies outside S. Conclusion: we must actually have G = S, as claimed. A Case Study: The Groups of Order 12. The following analysis of all groups of order 12 will draw upon almost everything discussed so far. 6.3.8 Example (Groups of Order 12). Classify all groups with |G| = 12 up to isomorphism. Identify all semidirect products in this family. Discussion: There are Sylow subgroups Hp for the prime divisors p = 2, 3, with |H2 | = 4 and |H3 | = 3; thus H3 Z3 while H2 could be either Z4 or Z2 Z2 (abelian). We = obviously have H2 H3 = (e) and |H2 H3 | = 4 3/|H2 H3 | = 12, so that H2 H3 = G. Now let Ep be the union of all conjugates of Hp . Then E2 E3 = (e) because all intersections Hp Hq are trivial when p = q. By the Sylow theorems we have #(Sylow 2-subgroups) = 1 (if H2 G), or 3 #(Sylow 3-subgroups) = 1 (if H3 G), or 4 If H3 is not normal the union of its conjugates has cardinality |E3 | = 4(3 - 1) + 1 = 9, since conjugates of H3 intersect only at the identity. In this situation, the other Sylow 37
subgroup H2 must be normal, otherwise E2 E3 would contain more than 12 points. (An argument of this sort cannot be made when H2 is non-normal because conjugates of H2 might overlap in subgroups of order 2.) The conclusion is At least one of the subgroups H2 , H3 must be normal in G and hence every G of order 12 can be realized as a semidirect product. We will employ the Chinese Remainder Theorem 6.1.26 in our analysis of these products. Case 1: Both H2 , H3 normal. Then G is the direct product H2 H3 and is abelian. The possible distinct isomorphism types are Group G(1) : Z3 Z4 Z12 = Group G(2) : Z3 (Z2 Z2 ) Z6 Z2 =
| | Case 2: Only H3 is normal. Then G is a semidirect product Z3 Z4 or Z3 (Z2 Z2 ). Case 2A: If H2 = Z4 , let us write both H2 and H3 = Z3 in additive notation. Then G is determined by some homomorphism
: Z4 (U3 , ) Aut(Z3 , +) = Since |U3 | = 2 the only possible assignments for the cyclic generator a = [1]4 in Z4 are (a) = id (in which case G is one of the abelian groups already listed), or (a) is the inversion map J([k]) = -[k], [k] Z3 . Then we we get a new group of order 12.
| Group G(3) . This group is the the semidirect product Z3 Z4 whose multiplication operation has been described in (28) of proposition 6.2.20:
([i]3 , [j]4 ) ([k]3 , []4 ) = (34) for all [i], [k] Z3 , [j], [] Z4 .
( [i] + (a)j ([k]) , [j] + [] ) = ( [i + (-1)j k]3 , [j + ]4 )
Note that (e) = (2 a) = (a)2 = id and (a), (3 a) = (a)3 are both equal to J. 6.3.9 Exercise. In terms of generators and relations, G(3) is generated by elements x, y which satisfy the relations x3 = e y4 = e yx = x2 y (or yxy -1 = x-1 )
| Verify this claim. Which elements in Z3 Z4 should be identified with x and y?
Case 2B: If H2 = Z2 Z2 let's write both H2 and H3 in multiplicative form H2 H3 = = {ui v j : i, j Z2 } = {e, u, v, uv} {e, a, a2 }
as in 6.3.4. Since Aut(H3 ) Aut(Z3 , +) (U3 , ) (Z2 , +), any nontrivial homomor= = = phism : H2 Aut(H3 ) will have a kernel ker() of index 2 in H2 . By changing our labeling of elements in H2 we may assume that is the map (e) = (u) = I (v) = (uv) = J (Inversion map J : ai a-i on H3 )
| The multiplication law for the resulting semidirect product G(4) = H3 H2 then takes the form
(ak , ui v j ) (a , ur v s ) (35)
= (ak (ui v j )(a ) , ui+r v j+s ) = (ak J j (a ) , ui+r v j+s ) = (ak+(-1) , ui+r v j+s )
j
for all exponents i, j, r, s Z2 and k, Z4 . This is a familiar group in disguise. 38
| Group G(4) . Up to isomorphism, this semidirect product Z3 (Z2 Z2 ) is the dihedral group D6
In fact, the element u H2 is central in G because (u) = I, and it generates a subgroup Z = u Z2 such that Z H3 = (e). Thus N = H3 Z is a subgroup isomorphic to = Z3 Z2 Z6 that is normal in G, with index |G/N | = 2. The element v in H2 = generates a copy of Z2 transverse to N . It is easy to check that the element = au is a cyclic generator for N and that = v, with o() = 2 satisfies the dihedral relation -1 = -1 . It follows that G(4) D6 as claimed. = | Case 3: Only H2 is normal. If H2 = Z4 then G = H2 H3 is determined by by some homomorphism : H3 = Z3 (U4 , ) = (Z2 , +). Since gcd(2, 3) = 1, must be trivial and we get nothing new. If H2 Z2 Z2 we again employ multiplicative notation for both H2 and H3 , as in = Case 2B. Important insight is achieved if we relabel elements of H2 = {ui v j : i, j Z2 } as e, x1 = u, x2 = v, x3 = uv Observe that xi xi+1 = xi+2 = xi-1 for 1 i 3 when subscripts are reckoned (mod 3), and that xi xi = e. (Thus for instance, x1 x2 = uv = x3 , x2 x3 = v uv = u = x1 etc.) Elements in Aut(H2 ) permute the xi leaving e fixed, and all permutations of [1, 3] are accounted for. 6.3.10 Exercise. Prove that every permutation of the integers [1, 3] yields an automorphism of Z2 Z2 = {e, x1 , x2 , x3 } such that (e) = e, (xi ) = x(i) for 1 i 3
Verify that the correspondence S3 Aut(H2 ) is bijective and an isomorphism of groups. Thus we obtain a natural identification of Aut(H2 ) with the permutation group S3 . A homomorphism : Z3 Aut(H2 ) S3 must carry the cyclic generator a H3 to a = permutation (a) = such that 3 = e. The case = e is uninteresting since it yields the trivial action, so we must assign (a) = (123) or (a) = (132) = (123)-1 . Both choices yield isomorphic semidirect products since they differ only in the way we label nontrivial elements in H2 , so we may as well take (a) = (123), which corresponds to the automorphism (a)(e) = e (a)(u) = v, (a)(v) = uv, (a)(uv) = u
| | of H2 . The new group G(5) = H2 H3 (Z2 Z2 )Z3 has the following multiplication = law
(36)
(ui v j , ak ) (ur v s , a ) = (ui v j (a)k (ur v s ) , ak+ )
for exponents i, j, r, s Z2 and k, Z4 . The role of the permutation (123) in this group law becomes clearer when we label elements in H2 as e, x1 , x2 , x3 . Then (a)xi = xi+1 and (a)k xi = xi+k if we reckon
39
subscripts (mod 3), and the multiplication law (36) takes the form (e, ak ) (e, a ) = (e, a ) (xi , a ) = = (xi , ak ) (e, a ) = = (xi , ak ) (xj , a ) = =
k
(e, ak+ ) (e (a)k (xi ) , ak+ ) (xi+k , ak+ ) (xi (a)k (e) , ak+ ) (xi , ak+ ) (xi (a)k (xj ) , ak+ ) (xi xj+k , ak+ )
To evaluate the last type of product we might need to make a 44 multiplication table for elements of H2 ; there is no simple algebraic formula m = m(i, j) for rewriting products xi xj in the form xm with 1 i, j, m 3. (The formula xi xi+1 = xi+2 only applies when j = i 1.) We have identified a new group.
| | Group G(5) : the semidirect product H2 H3 (Z2 Z2 )Z3 with multipli= cation law (36) is isomorphic to the group of even permutations A4 . It is also the group of orientation-preserving symmetries of the regular tetrahedron.
A regular tetrahedron centered at the origin T R3 is exceptional in that every permutation of its four vertices corresponds to an orthogonal linear transformation (rigid motion) in R3 that maps the tetrahedron to itself. It can be shown by convexity arguments that there are no other rigid-motion symmetries of this solid, so the full symmetry group of the tetrahedron is S4 . In this picture, the even permutations A4 correspond = to the subgroup of orientation-preserving symmetries, which are all rotations through axes passing through the center of T; the full group S4 includes some reflections across planes passing through the origin. Obviously |A4 | = 12. To see why the group described in (36) is isomorphic to A4 , regarded as the symmetry group of the tetrahedron T, consider the tetrahedron shown in Figure 6.8 with vertices labeled 1, 2, 3, 4. Any 22-cycle, such as (12)(34), corresponds to a 180 rotation about an axis passing through the midpoints of opposite edges. We have previously shown (Chapter 5) that the Klein Viergroup N = {e} {all three 2,2-cycles} is a normal subgroup in S4 isomorphic to Z2 Z2 . For each axis passing through a vertex and the center of the opposite face, we get a subgroup of order 3 (one of the four Sylow subgroups H3 ) consisting of 0 , 120, and 240 rotations about this axis. For example the subgroup {e, (123), (132)} corresponds to rotations about the axis passing through vertex 4 and the center of the opposite face. With these geometric descriptions in mind, it is not hard to see how to identify elements in our abstract model (36) with the correct geometric operations on T.
Figure 6.8. A regular tetrahedron centered at the origin. We show: rotation A1 by 120 about an axis through vertex 1, and B13 by 180 about an axis through midpoints of opposite edges.
6.3.11 Exercise. Explain why the five groups G(1) , . . . , G(5) of order 12 are pairwise non-isomorphic. We take up one last example to show that the Sylow theorems do not always yield a 40
definitive analysis, and that as the order of G increases we begin to encounter groups that are not semidirect products i.e. they arise as extensions e N G G/N e that do not split. 6.3.12 Example (Groups of order 8). These are all p-groups so the Sylow theorems are not much help; however by (3.2.5, 3.2.6) we know G has nontrivial center Z = Z(G), which can only have order 2, 4, 8. If |Z| = 8 the group is abelian and the possibilities are G = Z8 , Z4 Z2 , Z2 Z2 Z2 , in view of the following result. 6.3.13 Lemma. If G is abelian with |G| = 8 then G is isomorphic to Z8 , Z4 Z2 , or Z2 Z2 Z2 . Proof: If the maximum order of an element is o(x) = 8 then G Z8 . If the maximum = is 4, say for x = a, then A = a Z4 . Suppose there exists some b A such that / = o(b) = 2. Then B = b Z2 is transverse to A, A B = (e), AB = G, and since G is = abelian it is a direct product Z4 Z2 . = Otherwise we have o(b) = 4 for all b A and B Z4 , but now we must have / = |A B| = 2 (because |G| = |A| |B| = 16 if A B = (e)). The only subgroup of Z4 with order 2 is {[0]4 , [2]4 }, whose nontrivial element has order 2. That means a2 and b2 are the only elements of A and B of order 2, and we must have a2 = b2 , a-2 = a2 , b-2 = b2 , and A B = {e, a2 }. Now look at the powers of the element ab: we get e, ab, (ab)2 = a2 b2 = a2 a-2 = e. Since b A ab A we have found an element outside of A with / / order 2. Contradiction. This case cannot arise. If |Z| = 4 then G/Z Z2 . Since G/Z is cyclic and Z is central in G, G must be = abelian (why?). Contradiction. Thus |Z| cannot equal 4. 6.3.14 Exercise. If G is a finite group and Z a subgroup such that (i) Z is central in G (zg = gz for all z Z, g G), and (ii) G/Z is cyclic, prove that G must be abelian. Assuming |Z| = 2, let x be an element of highest order in G. We can only have o(x) = 2 or 4 (G is abelian if o(x) = 8). In the first case all elements y = e would have order 2 and G Z2 Z2 Z2 (see Exercises 6.2.28 - 29). Abelian G have already been = excluded, so we must have o(x) = 4 and N = x is isomorphic to Z4 ; the following elementary result shows that N is also normal in G. 6.3.15 Exercise. Let G be a group and H a subgroup of index |G/H| = 2. Prove that H is automatically normal in G. Hint: There are just two cosets, H and xH with x H. Show that xHx-1 = H. / We now have an extension e - N Z4 - G - Q = G/N Z2 - e = = and the real issue is whether it splits. If it does there is a subgroup Q = y Z2 transverse to G/N cosets and we have a = semidirect product N Q. We saw in 1.3.2 that Aut(Z4 , +) (U4 , ) (Z3 , +). The = = only possible homomorphisms : Z2 Aut(N ) must map the nontrivial element y Q to either (a) (y) = idN , in which case G is an abelian direct product Z4 Z2 , a possibility we have already excluded. (b) (y) = the inversion map, which takes [k]4 to -[k]4 = [4 - k]4 in Z4 . In this case: x4 = e, y 2 = e, yxy = yxy -1 = (y)(x) = x-1
Obviously this G is isomorphic to the the dihedral group D4 . 41
In the non-split case we claim that G is the group of unit quaternions Q8 = {1, i, j, k} in which the elements 1 are central, -i = (-1)i, . . . , -k = (-1)k and (-1)2 = 1 i2 = j 2 = k 2 = ijk = -1
from which we get other familiar relations such as ij = k = -ji, jk = i = -kj, . . .. Given Q8 it is easy to check that its center is Z(Q8 ) = {1, -1} and that N = j = {1, j, -1, -j} is a cyclic normal subgroup Z4 . But as you can easily check, the only elements g Q8 = such that g 2 = 1 are 1, which lie in N ; thus the extension Q8 of G/N Z2 by N Z4 = = cannot split. It remains to check by hand that up to an isomorphism Q8 is the only possibile nonsplit extension. Taking an x G that generates a cyclic normal subgroup N Z4 , let y = be any element lying outside N , so that G = x, y . Since our extension does not split we cannot have o(y) = 2, but y 2 e (mod N ) so y 2 {x, x2 , x3 = x-1 }. The first and last possibilities are excluded: if, say, y 2 = x then y 3 = xy = yx and G would be abelian. Hence our generators x, y satisfy o(x) = o(y) = 4 yN = x / y 2 = x2
Let us write "-1" for the element x2 = y 2 and "1" for the identity element in G. Then (-1)2 = 1 and -1 is central in G (because (-1)x = x2 x = x(-1), and likewise for y). Next observe that z = xy satisfies z 2 = -1. To see this, first note that xy N but / (xy)2 N . If xyxy = x then yxy = e and x = y 2 = x2 , which is impossible; likewise we get xyxy = x3 . So, we must have (xy)2 = x2 i.e. xyz = xy(xy) = -1. We conclude that G Q8 when we identify with 1, i, j, k with 1, x, y, z and -1 = x2 = y 2 , -i = = x3 , -j = y 3 , -k = z 3 . To summarize: the only groups of order 8 are Q8 , D4 , Z8 , Z4 Z2 , and Z2 Z2 Z2 . Deciding whether an extension e N G G/N e splits can be difficult. Even if N and G/N are cyclic, the extension need not split the group Q8 of quaternion units being one counterexample. Keeping in mind that G Zp for any group of prime order p > 1, we have now = identified all groups of order |G| 13 except for those of orders 9 and 10. You might try your hand at filling in these gaps. 6.3.16 Exercise. If |G| = 9 prove that (a) G is abelian (b) G is isomorphic to Z9 or Z3 Z3 . Explain why the groups in (b) are not isomorphic. 6.3.17 Exercise. Determine all groups of order |G| = 10
42
6.4 Some types of groups: simple, solvable, nilpotent.
A group is simple if it contains no proper normal subgroups, which is to say it has no proper quotients G/N ; by this definition, the trivial group is not simple. In view of Cauchy's theorem 4.3.5, the only simple finite abelian groups are (Zp , +) where p > 1 is prime. Noncommutative examples include the alternating groups An , consisting of all even permutations in the full permutation group Sn on n objects. The fact that An is simple for n 5 is of great importance in Galois' theory of equations; Sn itself is not simple because it always has An as a proper normal subgroup. Complementary to the simple groups we have the solvable and nilpotent groups. For any G, finite or not, two descending series of normal subgroups can be defined in terms of commutators [x, y] = xyx-1 y -1 . If A, B are subgroups we define the associated group [A, B] to be the subgroup generated by all commutators formed from elements of A and B: (37) [A, B] = xyx-1 y -1 : x A, y B
The commutator subgroup [G, G] obtained by taking A = B = G is of particular importance. First, it is a normal subgroup, but in fact it is a characteristic subgroup, which means it is invariant under all Aut(G). This follows because the image of a commutator [x, y] under any automorphism has the form (38) ([x, y]) = [(x), (y)] for all x, y G
and again is a commutator. Second, the quotient G/[G, G] is abelian because we are factoring out all relations associated with noncommutativity of G. In fact, the commutator group is the smallest normal subgroup such that G/N is abelian. 6.4.1 Exercise. Suppose S is a subset of a group G and let H = S be the subgroup it generates. If : G G is a homomorphism that maps S into itself, prove that leaves the generated subgroup invariant too i.e. (H) H. Hint: Recall that S is defined to be the smallest subgroup in G that contains S. It is also described by "building up from S" as S = {a1 a2 . . . ar : r < and ai S S -1 } i.e. S is the set of all "words" of finite length whose letters are of the form s or s-1 . The latter viewpoint might be congenial in proving that (H) H. 6.4.2 Exercise. If Aut(G) and x, y G, verify the identity (38) and then use it to prove that the commutator subgroup [G, G] is a characteristic subgroup in G. Hint: Use the previous exercise. 6.4.3 Exercise. If G is any group and N any normal subgroup, prove that G/N is abelian if and only if N [G, G]. Obviously G is abelian [G, G] is trivial. We now define the upper/lower derived series to be the descending series of subgroups shown in Figure 6.9. Both series begin with G followed by D1 (G) = D1 (G) = [G, G]. In order to understand these definitions it would be useful to verify that in the right hand series we have Dk+ (G) = Dk (D (G)). This is almost obvious (by induction on k); it is not true for the series on the left. The horizontal inclusions shown in Figure 6.9 follow by an easy inductive argument from the fact that A A [A , B] [A, B]. A recursive argument based on Exercises 6.4.1-2 shows that all subgroups in Figure 6.9 are normal, and in fact are characteristic subgroups in G. For example, we know that D1 (G) = [G, G] is characteristic by 6.4.2. The next derived group D2 (G) is generated by commutators [g, a] such that g G, a D1 (G); but an automorphism Aut(G) maps generators to generators in D2 (G) 43
G D1 (G) = [G, G] D2 (G) = [G, D1 (G)] D3 (G) = [G, D2 (G)] . . . Dk+1 (G) = [G, Dk (G)] . . . Upper Derived Series
=
G D1 (G) = [G, G] D2 (G) = [D1 (G), D1 (G)] D3 (G) = [D2 (G), D2 (G)] . . . Dk+1 (G) = [Dk (G), Dk (G)] . . . Lower Derived Series
Figure 6.9. The two derived series for G.
because ([g, a]) = [(g), (a)] [G, D1 (G)]. Hence by 6.4.1 the subgroup D2 (G) is invariant under . Similar arguments, which we omit, show that all the subgroups Dk (G) and Dk (G) are invariant under all automorphisms of G. 6.4.4 Exercise. Fill in the details needed to show (a) A A [A , B] [A, B] (b) The inclusions shown in Figure 6.9 are valid (c) The Dk (G) are all characteristic subgroups in G. (d) The Dk (G) are all characteristic subgroups in G. (e) Dk+ (G) = Dk (D (G)) for all k, 1. (f) If H G then Dk (H) Dk (G) and Dk (H) Dk (G). It is possible that one or both series stabilize after a finite number of steps, so that G D1 (G) . . . Dk (G) = Dk+1 (G) = . . . or G D1 (G) . . . Dk (G) = Dk+1 (G) = . . . , Clearly, once two successive groups are equal, say Dk (G) = Dk+1 (G), then all later subgroups are the same. Furthermore if G is finite the subgroups must "stabilize." When this happens the repeating stable subgroup need not be trivial; for example the alternating group An (n 5) has no proper normal subgroups and is nonabelian, so the lower derived series for the permutation group is Sn An = An = . . .. For infinite groups these descending series might not stabilize at all, as is true for the free group F2 on two generators. [One can, with some effort, prove that F2 is residually nilpotent in the sense that k=1 Dk (F2 ) = (e).] Two important classes of groups are defined by the properties of their derived series. 6.4.5 Definition. A group G is nilpotent if the upper derived series is eventually trivial, and G is solvable if the lower derived series becomes trivial in finitely many steps. Obviously (nilpotent) (solvable), but the converse fails. 6.4.6 Exercise. The affine group of the line G = Aff(2, R) consists of the operators T(a,b) : R R with a > 0, b R which form a group under composition. 44 T(a,b) (x) = ax + b
(a) Verify that G is a group and work out formulas for computing (i) T(a,b) -1 T(a ,b ) , and (ii) T(a,b) . (b) Verify that N = {T(1,b) : b R} is a normal subgroup isomorphic to R and that G/N R. Hint: The exponential map identifies (R, +) with the = multiplicative group of numbers a > 0. (c) Compute the derived group [G, G] and verify that G is solvable. (d) Show that the center Z(G) is trivial, and that G is not nilpotent. Note: This group is often referred to as the "ax + b group," for obvious reasons. 6.4.7 Exercise. The 3 3 Heisenberg triangular matrices 1 x G= 0 1 0 0 group can be described as the set of upper z y : x, y, z R 1
We shall label group elements by the symbol strings (x, y, z)
(a) In terms of these parameters compute the form of the product (x, y, z) (x , y , z ) = . . . of two group elements and the inverse (x, y, z)-1 = . . .. (b) Show that Z(G) = {(0, 0, z) : z R} and identify the commutator subgroup [G, G]. Then verify that G is nilpotent. We now list some basic combinatorial facts about these groups. The first is just an exercise in understanding the definitions. 6.4.8 Exercise. Show that any homomorphism : G G preserves the various derived subgroups: (i) (Dk (G)) = Dk ((G)) Dk (G ) (ii) (Dk (G)) = Dk ((G)) Dk (G ) for all k = 1, 2, . . .. Hint: Use Exercise 6.4.1 and induction on the index k. It follows almost immediately that all homomorphic images and all subgroups of nilpotent (or solvable) groups are again nilpotent (or solvable). [For quotients, apply Exercise 6.4.8. For any subgroup H we have [H, H] [G, G], and then inductively we get [H, Dk (H)] [G, Dk (G)] and [H, Dk (H)] [G, Dk (G)] for k = 1, 2, . . .. Thus Dr+1 (G) = (e) Dr+1 (H) = (e), and similarly for the lower derived series.] As a corollary we obtain a basic combinatorial result: 6.4.9 Lemma. If N G is a normal subgroup of a group G, giving us the sequence of homomorphisms id e - N - G - G/N - e , then G is solvable if and only if G/N and N are solvable. Proof: We have just discussed (), and conversely if : G G/N is the quotient homomorphism, solvability of G/N insures that Dn+1 (G/N ) = (e) for some n. Exercise 6.4.8 insures that (Dn+1 (G)) = Dn+1 (G/N ) = (e), and hence Dn+1 (G) N . Taking k so Dk+1 (N ) = (e) we get Dk+n+2 (G) = Dk+1 (Dn+1 (G)) Dk+1 (N ) = (e). Nilpotent groups do not share this property, see Exercise 6.4.6 where N R and G/N = = R are nilpotent (abelian) but G is not nilpotent. But for solvable groups there is an even stronger result which sheds light on how they are put together. 45
6.4.10 Lemma. A group G is solvable if there exist subgroups G = G0 G1 . . . Gr Gr+1 = (e) such that (i) Gj+1 is a normal subgroup in Gj for each j, (ii) each quotient Gj /Gj+1 is solvable. In particular G is solvable if the quotients are all abelian. Conversely, if G is solvable such sequences exist and the subgroups can be chosen so each quotient is abelian. Proof: The previous remarks show that we can work backward up the chain G D1 (G) D2 (G) . . ., successively verifying solvability of Gr , Gr-1 , . . .. Starting from the top, Exercise 6.4.3 shows that G/D1 (G) = G/[G, G] is abelian, but since D2 (G) = D1 (D1 (G)) = [D1 (G), D1 (G)] we see that D1 (G)/D2 (G) is abelian, etc. Nilpotent groups, unlike solvable groups, always have nontrivial centers. In fact, if Dr (G) = (e) and Dr+1 (G) = [G, Dr (G)] = (e), we see that Dr (G) is a nontrivial central subgroup in G. Central subgroups play a prominent role in the nilpotent analog of 6.4.10. 6.4.11 Lemma. A group G is nilpotent if there there is a subgroup N such that (i) N is central in G, and (ii) G/N is nilpotent. Proof: All subgroups of the center Z(Gj ) are normal in Gj , so G/N is a bona-fide group. Now observe that Dr+1 (G/N ) = (e) for some r and that Dr+1 (G/N ) = (Dr+1 (G)) under the quotient map . Thus Dr+1 (G) N = ker , and since N is central in G we get Dr+2 (G) = [G, Dr+1 (G)] [G, N ] = (e). Thus G is nilpotent. For other conditions leading to nilpotence of G see 6.4.13 below. One useful consequence of 6.4.11 is the fact that nonabelian finite groups always include sizeable nilpotent subgroups, namely their Sylow p-subgroups. 6.4.12 Corollary. If |G| = pk for some prime then G is nilpotent. Proof: If k = 1 then G Zp , so assume k > 1. Then the center Z(G) is nontrivial and = |G/Z(G)| = pr for some r < k. By induction G/Z(G) is nilpotent, and since Z(G) is central G must be nilpotent too. Nilpotent groups are the next best thing to being abelian; solvable groups also have some commutative aspects, but the connection is more tenuous. The pervasive role of "centers" in nilpotent groups is revealed by noting that there is a third canonical series of subgroups in any group G, the ascending central series Z(G) Z (2) Z (3) . . . defined as follows. Z (1) Z (2) Z (3) (39) Z (k+1) = = = . . . = . . . Z(G) (the center of G) {x G : xy yx (mod Z(G)), all y G} {x G : xy yx (mod Z (2) ), all y G} {x G : xy yx (mod Z (k) )}
The group Z (2) is often referred to as the second center of G; it is just the pullback to G of the center in the quotient group G/Z(G). (The center of the quotient need not be trivial.) Obviously, Z (k) Z (k+1) at every step, and if equality holds at the k th step 46
it holds at every later step. We note without proof the following characterization of nilpotent groups in terms of the ascending central series. 6.4.13 Theorem. A group G is nilpotent if and only if we have Z (k) = G for some k = 1, 2, . . .. Nilpotent groups are particularly amenable to study via inductive arguments. Indeed, 6.4.11 and 6.4.13 provide us with two inductive strategies: 1. Run up the ascending central series (e) Z(G) . . . hoping to prove some result by examining the abelian group Z(G) and the smaller nilpotent group G/Z(G). 2. Examine the descending derived series, G [G, G] . . . hoping to prove some result by examining the abelian group G/[G, G] and the smaller nilpotent group [G, G]. In this connection it is worth noting that any intermediate subgroup G H [G, G] is automatically nilpotent and normal in G, and G/H is abelian. 6.4.14 Example. The permutation groups S2 , S3 , S4 are solvable. For n 5 the alternating group An = { Sn : sgn() = +1} is simple and is the only proper normal subgroup in Sn . Hence Sn is not solvable for n 5. Discussion: Assume n 5. In 5.3.5 5.3.7 we indicated why An is the only proper normal subgroup in Sn , and showed that An is simple. The lower commutator series for Sn terminates prematurely, with Sn D1 (Sn ) = [Sn , Sn ] = An = D2 (Sn ) = D3 (Sn ) = . . . so Sn is not solvable. In fact, since the signature map sgn : Sn {1} is a homomorphism we get sgn(xyx-1 y -1 ) = sgn(x)sgn(y)sgn(x)-1 sgn(y)-1 = 1 for every commutator of elements in Sn . Thus [Sn , Sn ] An . But Sn is not abelian, so the the commutator subgroup [Sn , Sn ] is nontrivial and normal in Sn , and hence in An . Thus it is either all of An and D1 (Sn ) = [Sn , Sn ] = An as shown above. At the next step in the commutator sequence, An is nonabelian so D2 (Sn ) = [An , An ] is nontrivial, and it is a normal subgroup in An . As such, it must equal An . From here on the commutator sequence repeats. For n = 2, 3, 4 we know that S2 Z2 is abelian, hence solvable. In Example 5.4.1 = we showed that S3 /A3 Z2 and that A3 Z3 , so S3 is solvable by 6.4.10. In 5.4.4 we = = showed that S4 contains the abelian Klein Viergroup V4 Z2 Z2 as a normal sub= group, and that S4 /A4 Z2 , A4 /V4 Z3 . Applying 6.4.10 again we conclude that S4 = = is solvable. One goal of group theory has been to classify the finite simple groups up to isomorphism. Nilpotent and solvable groups are generally far from simple, owing to the presence of various series of normal subgroups. It is easy to see that the only simple solvable groups are abelian, and they are the cyclic groups (Zp , +) where p > 1 is a prime. Here we can only mention a remarkable theorem proved in the 1970's that opened the way for the successful classification of all finite simple groups in the 1980's. 6.4.15 Theorem. If G is a finite group of odd order, then G is solvable. Hence all nonabelian finite simple groups have order divisible by 2. The proof ran some 250 pages and occupied an entire issue of Pacific Journal of Mathematics. Note that the only abelian simple groups are (Zp , +) for primes p > 1; these have odd order p, except for p = 2.
47
The structure of non-simple groups can be quite complicated, even though they are in some sense assembled by combining simple groups. Nevertheless, certain structural features can be identified by recalling that the product set N = N1 N2 formed from two normal subgroups is again a normal subgroup. First notice that if both groups are solvable, so is the product. In fact, by the Second Isomorphism Theorem for quotients 3.3.14 we have a sequence of homomorphisms
id e - N1 - N - N/N1 N2 /N1 N2 - e =
and the middle group is solvable if the end groups are (Lemma 6.4.9). Thus a group G always contains a unique largest solvable normal subgroup R, which is sometimes referred to as the solvable radical of G. The quotient G/R contains no normal solvable subgroups at all, although there will certainly be non-normal abelian and solvable subgroups in it, for instance cyclic subgroups. 6.4.16 Exercise. For any n 3 determine the center Z(G) of the dihedral group G = Dn . Hint: The answer will depend on whether n is even or odd. 6.4.17 Exercise. Compute the commutator subgroup [G, G] for the dihedral group G = Dn , n 3. Is Dn nilpotent for any n? Solvable? 6.4.18 Exercise. In the cases where the center of Dn is nontirival, does the extension e Z(Dn ) Dn Dn /Z(Dn ) e split i.e. is Dn a semidirect product with its center as the normal subgroup? 6.4.19 Exercise. Compute the commutator subgroup [G, G] of the quaternion group G = Q8 of Example 6.3.10. Is Q8 nilpotent? Solvable?
48

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Notes: F.P. Greenleaf c 2000 - 2010v43-f10sn.tex, version 11/11/10Algebra I: Chapter 5. Permutation Groups 5.1 The Structure of a Permutation.The permutation group Sn is the collection of all bijective maps : X X of the set X = cfw_1, 2, . . . , n, wit

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Notes: F.P. Greenleaf, 2000-2010v43-f10trgps.tex, version 11/5/10Algebra I: Chapter 4. Transformation Groups 4.1 Actions of a Group G on a Space X.Let G be a group and X a set. 4.1.1 Definition. A group action is a map : G X X that assigns to each pair

MO St. Louis - MARKETING - 110

The G.M. CaseClass NotesTHE G.M. CASE SYNOPSISSYNOPSIS OF THE CASE 1. HENRY FORD STRATEGYA. LOW - COST B. MASS PRODUCTION C. DEVELOPED MASS MARKET - ONE MODEL "T" D. LIMITED PRODUCT DIFFERENTATION E. MFG. DISTINCTIVE COMPETENCE STANDARDIZATION OF QUAL

University of Phoenix - CRT - 205

Source 1 Title and Citation: Drinking Impairs Youth Development Drinking Impairs Youth Development. U.S. Department of Health and Human Services. Opposing Viewpoints: Alcohol. Ed. Andrea C. Nakaya. Detroit: Greenhaven Press, 2008. Viewpoint essay. Retrie

UC Davis - ENG - 101

AQUA-CSPConcentrating Solar Power for Seawater DesalinationFinal Reportby German Aerospace Center (DLR) Institute of Technical Thermodynamics Section Systems Analysis and Technology AssessmentStudy commissioned by Federal Ministry for the Environment,

NJIT - PHYSICS - 121

PHYS 1443-501, Spring 2002, 2nd -Term Exam, Wednesday, Apr. 10, 2002Name: ID: [ 1 20 points] A uniform rod of length 150 cm and mass 900g is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane, as shown

NJIT - PHYSICS - 121

billing (cab4763) hw 2 opyrchal (121104) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points Consider three charges arranged as shown.

NJIT - PHYSICS - 121

billing (cab4763) hw 3 opyrchal (121104) This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points An electric field of magnitude 2210 N/C is

NJIT - PHYSICS - 121

billing (cab4763) hw 4 opyrchal (121104) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 WITHDRAWN 0.0 points1Since the electrostatic field in a conducting

NJIT - PHYSICS - 121

billing (cab4763) hw 5 opyrchal (121104) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A 27 m length of coaxial cable has a solid

NJIT - PHYSICS - 121

billing (cab4763) hw 6 opyrchal (121104) This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The current in a wire decreases with time according to t

NJIT - PHYSICS - 121

billing (cab4763) hw 7 opyrchal (121104) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A battery with an emf of 6.1 V and interna

NJIT - PHYSICS - 121

billing (cab4763) hw 8 opyrchal (121104) This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points 27.9 V 22.9 = (I3 - I1 ) r2 + I3 R = -I1 r

NJIT - PHYSICS - 121

billing (cab4763) hw 9 opyrchal (121104) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points An electron in a vacuum is first accelerated by a voltag

NJIT - CHEM 126 - 126

Equilibrium CalculationsInitial [HI]:0.00200mol = 4.00 x10-4 M 5.00 LInitial: Change:4x10-4 -2x0 +x x0 +x xReger et al p 604Eqm: 4x10-4 -2xGiven [I2] eqm = 3.8 x 10-5 M = xEquilibrium concentrations: [I2] = 3.8 x 10-5 M [H2] = 3.8 x 10-5 M [HI]

NJIT - CHEM 126 - 126

Equilibrium ConcentrationHint, finish filling-in the following table, then use the equilibrium concentrations in terms of x in the equilibrium expression and solve for x.Concentration and Le Chatelier's PrincipleConcentration Le Chatlier's Principle st

NJIT - CHEM 126 - 126

COMBUSTION TRAINAlso,Mg(ClO4)2 Absorbs: CO2 H2OPerchlorate ion: ClO4-A 2.074 g sample of a compound containing only carbon, hydrogen, and oxygen burns in excess oxygen to produce 3.80 g of CO2 and 1.04 g of H2O. Find moles and masses of each element an

NJIT - CHEM 126 - 126

CHAPTER 17 SUPPLEMENTS IHeat capacity calculationRemember T always equals Tfinal - TinitialSpecific heat definition and exampleSpecific heat 2nd exampleSpecific heat 3rd exampleLaw of Dulong et PetitThe specific heat is the heat capacity per gram.

NJIT - CHEM 126 - 126

CHAPTER 17 SUPPLEMENTS IIExamples of spontaneous processesUse your every day experience and the following examples to help you.Spontaneous processes and change in energySpontaneous processes and change in entropyEnthalpy versus entropyConditions whi

NJIT - CHEM 126 - 126

Chem 126 Common 2 Spring March 11, 2011NONE OF THE MULTIPLE CHOICE PROBLEMS REQUIRES EXTENSIVE OR TIME COMSUMING CALCULATIONS. IF YOUR METHOD REQUIRES EXTENSTIVE CALCULATIONS IT IS EITHER WRONG OR HARDER THAN WHAT IS REQUIRED. 1. The pH of a nitric acid,

NJIT - CHEMISTRY - 126

FE2Acids and Bases I. Arrhenius Acids and Bases:HNO3 HClO4 (perchloric acid) HCl HBr HI H2SO4 (sulfuric acid) HF CH3COOH PropertiesAn acid-substance that produces [H+] in H2O. Taste sour (don't tasteany that aren't foods) Neutralize bases React with m

NJIT - CHEMISTRY - 126

TITRATIONSEquivalence Point Acid-Base Indicator End PointMcQuarrieTITRATIONSMcQuarrieTITRATION OF STRONG ACID WITH STRONG BASEMcQuarrieFE11TITRATION OF WEAK ACID WITH STRONG BASEpH at midpoint? B- 0 = pK a pH = pK a + log [ HB ]0 Note, the equi

NJIT - PHYSICS - 121

ELECTROCHEMISTRY (GALVANIC OR VOLTAIC CELLS)These are named for Allesandro Volta and Luigi Galvani, two Italian physicists of the 18th century.This setup is like popping an inflated balloon-no useful work is done. It is possible to use Zn and Cu2+ to do

NJIT - PHYSICS - 121

GALVANIC CELLELECTROLYTIC CELL +-+AnodeCathode Cu deposits on electrodeCathode Zn deposits on electrodeAnode Required voltage: > 1.10 V OvervoltageRegerFE6ELECTROLYSIS charge = current x time q = It q (in coulombs,C ) I (in amperes, A = C/s) t (

NJIT - PHYSICS - 121

BATTERIES Galvanic cells are batteries (not practical) Example: Oxidation Reduction Why not good Battery: Fragile Low current Low energy density Inefficient Conversion to more useful battery includes: No H2 source-higher voltage, less complicated, and che

NJIT - PHYSICS - 121

+1 +2 Common oxidation states based on the periodic table. There are exceptions, a better procedure will be shown. +3 +4 -3 -2 -1OXIDATION STATES (REVIEW) Use Periodic Table to get the oxidation state. The composition is found from the conservation of ch

NJIT - PHYSICS - 121

ELECTROCHEMISTRY (LARGELY REVIEW) DISPLACEMENT 0XIDATION-REDUCTION REACTIONS Alkali Metals MOST ACTIVE GROUP OF METALSLow density (g/mol): Li 0.53, Na 0.97, K 0.89, Rb 1.53, Cs 1.93 Soft: Sodium cut with a knife (Potassium even softer)Small first ioniza

NJIT - PHYSICS - 121

CHAPTER 18 SUPPLEMENTS IPeriodic Table and oxidation stateComposition of Ionic CompoundsOxidation State RulesExamples of the application of the rules given below. If 2 rules are in conflict, the rule with the smaller number wins or takes precedence. T

University of Toronto - MAT - 102

1.13. Let x A. Then x = 2k - 1 for some k Z. Let k1 = k - 1. We then have 2k1 + 1 B, but also 2k1 + 1 = 2(k - 1) + 1 = 2k - 1 = x, so x B. Hence A B. Conversely, let x B so that x = 2m + 1 for some m Z and let m1 = 2m - 1. Then 2m1 - 1 A and x = 2m + 1 =

University of Toronto - MAT - 102

1.7. Let x = 1, y = -1. Then -1/x = -1 < 1 = -1/y. The statement is true if we add the condition that y > 0. For if x > y > 0, then x/y > y/y = 1, so 1/y > 1/x. (Note that this depends crucially on the fact that x and y are positive.) But this yields 0 >

University of Toronto - MAT - 102

1.7. The domain of the absolute value function is all of R since R = (-, 0] [0, ). The image I is S = cfw_x R : x 0. If x S, then |x| = x, so x I. If x S, then x < 0, so |y| = x for any y R since the definition of the absolute value / function ensures tha

University of Toronto - MAT - 102

2.3. In symbols, the sentence is "(a R)(x R)P (a, x) Q(x)." This is false since P (0, 1) is true, but Q(1) is false. One way to change the quantifiers to make a true statement is "(a R - cfw_0)(x R)P (a, x) Q(x)." Another way is "(a R)(x R)P (a, x) Q(x)."

University of Toronto - MAT - 102

2.5. (a) We claim x = (y - b)/m is a solution. We have m y-b + b = (y - b) + b = y, mso this is indeed a solution to y=mx+b as claimed. To show uniqueness, suppose x1 and x2 are solutions. Then mx1 + b = y = mx2 + b mx1 = mx2 x1 = x2 . In other words, al

University of Toronto - MAT - 102

3.1. Let P (n) be "n < 100". 3.2. Let Q(n) be P (n). By induction, Q(n) is true for all n N, so P (n) is false for all n N. 3.6. This is false. Let P (n) be "n > 1." 3.11. We use induction on n. If n = 1, then a set with n elements has precisely 2 = 21 su

University of Toronto - MAT - 102

3.33. The largest length for a subinterval as described in the exercise is n - 1 (consider [1, n] itself) and the smallest is 0 (obtained by the intervals of the form [i, i] , 1 i n). Fixing i N, 0 i n - 1, we wish to count the number of subintervals of l

University of Toronto - MAT - 102

4.5. There is such a bijection if and only if A contains at least two elements. If A contains only one element or is empty, then the only function from A to A is the identity function. If A does contain at least two elements, then let x, y A, x = y. Defin

University of Toronto - MAT - 102

4.12 (a) False. One counterexample is f (x) = e-x . (b) False. Take f (x) = 0. (Any constant function will work.) (c) False. Consider x if 0 = x = 1, 1 if x = 0, f (x) = 0 if x = 1. (d) True. If |f (x)| M for every x R, then M + 1 is not in the image of f

University of Toronto - MAT - 102

4.26 Suppose f (x) = f (y). Then |f (x) - f (y)| = 0, so c |x - y| 0. Since c > 0 and > 0, this implies that |x - y| = 0 and hence x = y. Thus f is injective. 4.32 We have f (-x) = -(-x) = x and g(y -1 ) = (y -1 )-1 = y for every x F and y F - cfw_0, so f

University of Toronto - MAT - 102

4.33 (a) Let f : A B, g : B C be injections. Suppose (g f )(x) = (g f )(y). This means g(f (x) = g(f (y). Since g is injective, this implies that f (x) = f (y). But f is also injective, so x = y. It follows that g f is injective. (b) Let f : A B, g : B C

University of Toronto - MAT - 102

Homework 12 - MATH 310 0101 - Summer I 2010 - Due by Thursday, June 241. Prove that if A is countable, then Ak is countable for every k N. (Hint: Use induction. You will want to use the fact that Ak+1 = Ak A ; you must prove this in your solution.) 2. Pr

University of Toronto - MAT - 102

1 First we outline the proof that Ak+1 = Ak A . We define a map (a1 , a2 , . . . , ak+1 ) (a1 , a2 , . . . , ak ), ak+1 ). This map is easily shown to be a bijection from Ak+1 to Ak A. Now we prove the main result by induction on k. For the base case, we

University of Toronto - MAT - 102

Homework 13 - MATH 310 0101 - Summer I 2010 - Due by Thursday, June 25For each of the following relations, determine whether it is reflexive, symmetric, and/or transitive. In each case either prove it satisfies the property or display a counterexample. U

University of Toronto - MAT - 102

1. Reflexive, symmetric, and transitive. Equivalence relation. 2. Reflexive and symmetric, but not transitive. Not an equivalence relation. 3. Transitive, but neither reflexive nor symmetric. Not an equivalence relation. 4. Symmetric, but neither reflexiv

University of Toronto - MAT - 102

Homework 14 - MATH 310 0101 - Summer I 2010 - Due by Wednesday, June 301. Exercise 7.6 2. Exercise 7.8 3. Let A and B be partial orders on A and B respectively. Define a relation on A B by (a, b) (a , b ) if and only if either a A a and a = a or a = a an

University of Toronto - MAT - 102

Homework 15 - MATH 310 0101 - Summer I 2010 - Due by Wednesday, June 301. Let S R. Prove that if S has a minimum element , then S has an infimum with inf S = . 2. Prove that inf(a, b) = a. 3. Exercise 13.8 4. Exercise 13.19 5. Exercise 13.20 6. Exercise

University of Toronto - MAT - 102

1. By the definition of a minimum, x for every x S, so is a lower bound for S. If is an another lower bound for S, then, in particular, since S. Hence must be the greatest lower bound of S. In other words, = inf S. 2. a is a lower bound for (a, b) simply

University of Toronto - MAT - 102

13.12 This is false. Let S = cfw_0, 1 and set xn = 1, yn = 0. 13.23 Since sup A a and sup B b for every a A and b B, sup A+sup B a+b for every a+b C. Thus sup A+sup B is an upper bound for C. By Proposition 13.15, there are sequences an in A and bn in B s

University of Toronto - MAT - 102

Homework 17 - MATH 310 0101 - Summer I 2010 - Due by Thursday, July 81. Exercise 14.1 2. Exercise 14.2 3. Exercise 14.29 4. Define f : R R by f (x) = -1 if x < 0, 1 if x 0.Show that f is not continuous at 0. Do this in two ways: (a) Use the " - " defini

University of Toronto - MAT - 102

1. n is an unbounded sequence with no bounded, and hence no convergent, subsequence. an defined by an = n if n is odd 0 if n is evenis an unbounded sequence that has 0 as a convergent subsequence. 2. (a) Let an = (b) Let an =n i=1 n i=11/n. n.(c) Let

University of Toronto - MAT - 102

Homework 18 - MATH 310 0101 - Summer I 2010 - Due by Thursday, July 81. Show that the discrete metric is a metric. 2. Show that (a, b) is open in R. 3. Show that the union of open sets is open. 4. Show that a finite subset of a metric space is closed. 5.

University of Toronto - MAT - 102

1. We simply verify the properties of a metric. Positive definite: This is just a trivial observation from the definition. Symmetric: Obviously d(x, x) = d(x, x). If x = y, then d(x, y) = 1 = d(y, x). Triangle Inequality: If x = z, then d(x, z) = 0 = 0+0

University of Toronto - MAT - 102

MAT102S - Introduction to Mathematical Proofs - UTM - Spring 2010 Solutions to Selected Problems from Problem Set DFor the first two questions (about fields) see solutions to Quiz #2. 2.2. Take a = 0 and b = 1. The statement claims that there are integer

University of Toronto - MAT - 102

University of Toronto - MAT - 102

University of Toronto - MAT - 102

MAT102S - Introduction to Mathematical Proofs - UTM Problem Set A - Spring 2010Here are the Exercises assigned: 1.2. Fill in the blanks. The equation x2 +bx+c = 0 has exactly one solution when and it has no solutions when . 1.5. (-) Consider the Celsius

University of Toronto - MAT - 102

University of Toronto - MAT - 102

University of Toronto - MAT - 102

University of Toronto - MAT - 102

University of Toronto - MAT - 102

MAT102S - Introduction to Mathematical Proofs - Spring 2010 - UTM Problem Set E - TO BE SUBMITTED TO YOUR TADue:Monday, February 8, in tutorials. This assignment must be submitted to your TA at the beginning of the tutorial on the above date. Marking s