375.6
1.
Class MATH 6: Selected Answers
Gallian page 52 6.
Solution:
Look at a non-abelian group, say our old friend D3 . Notice that a,1 ba 6= b ba 6= ab: In D3 , if a and b are any two re ections, then ba 6= ab.
r120 r240 a b c
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r0
r0 r0 r120 r240 a b c
r120 r120 r240 r0 c a b
r240 a b c r240 a b c r0 b ca r120 c a b b r0 r240 r120 c r120 r0 r240 a r240 r120 r0
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Gallian page 52 8. Prove that a,1 ban = a,1 bn a. Solution: We will rst show that this is true for the non-negative integers by induction. Base case: n = 0. Then a,1 ba0 = e and we compare this to a,1 b0 a = a,1 ea = e, so the the induction starts. Inductive step: Assume a,1ban = a,1 bn a and now show a,1 ban+1 = a,1 bn+1 a. But a,1 ban+1 = a,1 ban a,1 ba = a,1 bn a a,1 ba = a,1 bn eba = a,1 bn+1 a: So the result is true for all non-negative integers. Now consider ,n where n 2 Z+ . Then using the fact that the inverse of a product is the product of the inverses in reverse order, a,1 ba,n = a,1 ban ,1 = a,1 bn a ,1 = a,1 bn ,1 a,1 ,1 = a,1 b,n a: This completes the proof. Gallian page 53 24. Prove that every Cayley table is a Latin square. Solution: Assume not. Assume that in there is an element a 2 G so that in the a-row of the table, the same element, say x, appears twice. This means that there are two distinct elements, say s; t 2 G such that as = x and at = x. But then as = at and by left cancellation, s = t. This contradicts that s and t are distinct. So the same element cannot appear twice in any row. A similar argument works for columns and uses right cancellation. Gallian page 53 26. Prove that if ab2 = a2 b2 in a group G, then ab = ba. Solution: Finally, an easy one. Just write it out. ab2 = a2 b2 abab = aabb bab = abb ba = ab where we have used left and right cancellation in the last two steps. Let H n denote the set of n n symmetric matrices. That is, H n = fA 2 Mnn j AT = Ag; where AT denotes the transpose of A. Show that H n is a subgroup of Mnn , the group of all n n matrices under addition. Solution: Check closure and invervses. Closure: Let A; B 2 H n. Then A = AT and B = B T . Show A + B is symmetric. A + B T = AT + B T = A + B: So A + B is symmetric. Inverses: Remember the group operation is addition. Then if A is symmetric, we must show ,A is symmetric. Now AT = A and one can pull scalars out of the transpose operation, so ,AT = ,AT = ,A = ,A: So ,A is symmetric and H n is a subgroup of Mnn
6.
a
The Heisenberg Group Heisenberg was a Nobel prize winner in physics is the set of 3 3 matrices de ned by: 80 1 a b 1 9 = H = :@ 0 1 c A a; b; c 2 R; : 001 Show that H is an subgroup of GL3, the group of 3 3 matrices under multiplication. Solution: Closure: Let 01 a b1 01 d e 1 A = @0 1 cA; B = @0 1 f A: 001 001 Then 0 1 a + d b + af + e 1 AB = @ 0 1 c + f A: 00 1 Note that AB has the correct form to be in H . Inverses: From the product AB above, you can see what the inverse has to be. If we want B to be the inverse, then
01 AB = I = @ 0
00 1 0A: 001
1
So comparing to the earlier calculation, we need d = ,a, f = ,c, and e = ,b + ac. Alternately, you could get theinverse by the usual reduction process. In either case:
0 1 ,a ,b + ac 1 A,1 = @ 0 1 ,c A :
00 1
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Note that this matrix has the correct form to be in H . In H , nd the order of the element 01 0 11 @0 1 0A: 001 Solution: From the calculation of AB above, it follows that for any n 2 Z+
01 @0
01n 10n 1 0A = @0 1 0A: 001 001
1
0
1
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So the order is 1. c What jH is j? Solution: jH j = 1 since it contains at least the in nite number of matrices from the previous part. Find j8j in Z10 and j8j in U 9. Solution: j8j = 5 in Z10 since 5 8 = 40 0 mod 10 and j8j = 2 in U 9 since 82 = 64 1 mod 9. a Gallian page 65 4. Prove that any element a and its inverse a,1 have the same order. Solution: If both elements have in nite order then we are done. So assume that at least one of a or a,1 has nite order. Suppose a has order m. Then
am = e am,1 = e,1 = e:
So a,1 has nite order. And similarly, if a,1 has nite order, so does a just reverse the arrow above. Now we check to see if they are the same order. The potential problem is that both could have nite order, say 4 and 8, but not the same order. So assume jaj = m, so m is the smallest positive integer such
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that am = e. And assume ja,1 j = n, so n is the smallest positive integer such that a,1 n = e. However, we just saw that am = e am ,1 = e; so this means that n m since n is the smallest power of a,1 to produce e. Of course, a,1 n = e a,1 n ,1 = an = e,1 = e; so now m n. Therefore m = n. b Prove that the number of elements x in a group G such that x3 = e is odd. Solution: Clearly e3 = e. Now if x 6= e then x2 = x,1 because xx2 = x3 = e. This also means that x2 6= e otherwise we would have x2 = x,1 = e x = e. But from part a, both x and x,1 have the same order, namely 3. That is, the elements of order 3 come in pairs of the form x and x2 = x,1 . So the the number of elements of order 3 is even. But we also have that e3 = e, so the total number of elements satisfying the condition is odd. a Gallian page 68 38 a and c. Solution: jU 3j = 2, jU 4j = 2, jU 12j = 4. jU 4j = 2, jU 5j = 4, jU 12j = 8. b Conjecture: jU mj jU nj = jU mnj, at least if gcdm; n = 1. This problem combines linear algebra, trigonometry, and abstract algebra. Great! For any real number , let cos , sin R = sin : cos
a
Show that R 2 SL2; R. Solution: det R = cos2 + sin2 = 1 R 2 SL2; R: Show that R R = R + .
Solution:
b
R R = sin
cos sin cos cos , sin cos sin , sin = sin cos + cos sin , os c...