375 Math Week 9
9.1 Normal Subgroups
We have seen the two pieces of the folowing result earlier in the course. THEOREM 1 Let H be a subgroup of G and let g 2 G be a xed element. a The set gHg,1 is a subgroup of G. b jH j = jgHg,1j. Make sure that you understand this. Try proving this right now before reading the proof below. Begin by using the two-step test to prove the rst part and then show that the mapping : H ! gHg ,1 is an isomorphism which proves the two sets have the same number of elements. PROOF A few weeks ago we saw that if : G1 ! G2 is a group homomorphism and H G1 , then H G2 . So for a xed g 2 G, de ne : G ! G by x = gxg ,1 . We will show that is an isomorphism and this will prove both parts of the theorem. is a homomorphism because a b = gag ,1gabg ,1 = g abg ,1 = ab: is injective because by cancellation, a = b gag ,1 = gbg ,1 a = b: is surjective because for any x 2 G codomain, g ,1 xg 2 G domain since G is a group closed. But g ,1 xg = g g ,1xg g ,1 = x: Then observe that part a of the theorem is true because H = gHg,1 and part b is true because is an isomorphism: jH j = j H j = jgHg,1j. We now single out certain subgroups H for special attention:
1
2
DEFINITION 2
Math 375
Let H be a subgroup of a group G. Then H is normal in G if gH = Hg for all g 2 G left and right cosets are equal. This is denoted by writing H G. At this point, it is not entirely clear why you would want to look at normal subgroups. But we will see that if H G, then the set of all left cosets of H is itself a group in a natural way. By next class, we should be able to prove this. For the moment, let's review what we done in lab and previous classes where we looked at left and right cosets. We saw that: An Sn , S2 is not normal in S3, and J Q8 . The next theorem gives us a criterion that makes it possible to show a H G without having to actually compute gHg ,1 for every g in G. THEOREM 3 Let H be a subgroup of G. Then the following three conditions are equivalent. i H is normal in G, that is, gH = Hg for all g 2 G; ii for all g 2 G, we have H = gHg ,1; iii ghg ,1 2 H for every g 2 G and every h 2 H , that is, gHg ,1 H . PROOF Notice that i ii was done as part of the coset property theorem. To do ii iii is easy: Since gHg ,1 = H , then we must have ,1 2 H for all g 2 G and all h 2 H . ghg Finally, for iii ii: This is really just the fact that the map in Theorem 1 is onto. ghg ,1 2 H for all g 2 G and h 2 H implies gHg,1 H . Now we need to show that for any g 2 G, H gHg,1. That is, given any g 2 G, we must show that any h 2 H can be written in the form h = gh0 g ,1 , for some h0 2 H . In particular, let x = g ,1 . Now set h0 = xhx,1 . Then by assumption iii, we know that h0 2 H . So h = gg,1hgg,1 = gxhx,1g = gh0g,1 2 gHg,1; which is what we wanted to show. EXAMPLE a Use condition iii to show that SLR; n GLR; n. b Along the same lines: Let GL+ R; be the the set of matrices with positive determinant. This is a subgroup of GLR; n. You might check that. Show that GL+ R; n GLR; n. THEOREM 4 Let H be a subgroup of G such that G : H = 2. Then H G.
EXAMPLE 2
PROOF
THEOREM 5
The two left cosets are H and G , H . But so are the right. Now give another proof that GL+R; n GLR; n. C G G. Further, if H is a subgroup of C G, then H G.
9.1 Normal Subgroups
PROOF
COROLLARY 6
3
Let H be a subgroup of C G. Then 8g 2 G and 8h 2 H , we have ghg,1 = hgg,1 = h 2 H . So by condition of iii the Theorem 3, H G. Let H be a subgroup of an abelian group G. Then H G. Note C G = G. Let H be a subgroup of a cyclic group G. Then H G:
PROOF
COROLLARY 7
PROOF
G is cyclic so it is abelian.
There are other simple criteria for normality which depend more on the subgroup H , than the group G. Here's one that is sometimes useful. If H is a subgroup of G and no other subgroup of G has the same order as H , then H is normal. We know that for any g 2 G that gHg ,1 is a subgroup of G and that jH j = jgHg,1j. But there are no subgroups of the order of H in G except H itself. So we must have gHg ,1 = H for all g . Let G = S3 Z5 . Then clearly G is not abelian since S3 is not. Let
COROLLARY 8
PROOF
EXAMPLE 3
H = 1; 1 = f1; 1; 1; 2; 1; 3; 1; 4; 1; 0g:
Show that H G. SOLUTION Notice that G : H = 6 6= 2 and G is not abelian so none of the obvious results on normality apply. However if we can show that H is the only subgroup of G of order 5, then H G. We can do this by using a proof by contradiction. Suppose some other subgroup K also had order 5. Then K is cyclic why?, thus K = ; n why? where 2 S3 and n 2 Z5 . We know that 5 = j ; n j = lcmj j; jnj: What are the possible orders of ? What are the possible orders of n? Show that the only way the lcm to be 5 is for j j = 1 and jnj = 5. Thus = 1 and n can be 1, 2, 3, or 4 why?. In any of these cases K = 1; n = H . Thus H G. Can you also do this same problem above by showing H is a subgroup of C G? For another illustration of this last corollary, see p. 187 39.
EXAMPLE 4 EXAMPLE 5
4
DEFINITION 9
Math 375
Let H be a subgroup of G. Let
N H = fg 2 G j gHg,1 = H g: N H is called the normalizer of H in G.
EXAMPLE a b
LEMMA 10
Find the normalizer of another subgroup of a nonabelian group. The normalizer of a subgroup is important because: a N H is a subgroup of G; b H N H ; c H N H . Use the one-step test: Let a; b 2 N H . This means that aHa,1 = H and bHb,1 = H . The latter implies that H = b,1 Hb. We must show ab,1 2 N H , that is, ab,1H ab,1,1 = H . But ab,1 H ...