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Math 545 Midterm Solutions Math 545 Midterm Solutions May 16, 2019 Hi everyone, please enjoy these solutions to the midterm. 1. All of these can be found in the book. 2. (a) (i) U(10) = f1; 3; 7; 9g are the integers mod 10 between 0 and 10 that are relatively prime to 10. A Cayley table is as follows: 1 3 7 9 1 1 3 7 9 3 3 9 1 7 7 7 1 9 3 9 9 7 3 1 (ii) It is cyclic, U(10) =< 3 >, it is also generated by 7 as well. (b) (i) Here is the Cayley table: e (12) (13) (23) (123) (132) e e (12) (13) (23) (123) (132) (12) (12) e (132) (123) (23) (13) (13) (13) (123) e (132) (12) (23) (23) (23) (132) (123) e (13) (12) (123) (123) (13) (23) (12) (132) e (132) (132) (23) (12) (13) e (123) (ii) The center is trivial. Look at the Cayley table: for each element σ in S3, there is another element τ for which στ 6= τσ. (c) (i) Not a subgroup: No matrix of this form (except the zero matrix) has an (additive) inverse. (ii) There is no identity, and, it is not closed under addition. (iii) This is a subgroup. Consider the one-step subgroup test, and let σ; τ 2 H. Then σ(5) = τ(5) = 5, which means that τ −1(5) = 5 as well. So, the permutation στ −1(5) = 5; so στ −1 2 H. ∗ a 3. (a) Suppose x is a generator of Q . Then x = b for integers a and b. Consider the prime factors of a and b: there must be only finitely many. Since the primes 1 are infinite, choose a prime number p not appearing in either factorization. This a number p is not in < x >, since powers of x = b must only involve the prime ∗ factors in either a or b. Thus, Q is not cyclic. (b) (i) The identity element in this group is the identity map i(x) = x. −1 x−2 −1 −1 (ii) Consider the function f (x) = 3 . Clearly f ◦ f = f ◦ f = i, so f −1 2 C(f). But it is not hard to see that f −1 does not commute with the function g(x) = 2x: 2x − 2 x − 2 2 f −1(g(x)) = f −1(2x) = ; g(f −1(x)) = g( ) = (x − 2): 3 3 3 So f −1 2 C(f), but we've shown that f −1 2= Z(G), so that in general Z(G) 6= C(a) for some a 2 G. (c) (i) (135)(24678) (ii) 15 (iii) It is even: it is the product of two cycles of odd length, each of which is even, and so the product of these is even as well. (d) Since H is closed under the operation, and a 2 H, we have fa; a2;:::g ⊆ H. 2 Since H is finite, we must have fa; a ;:::g is finite, and so there exists i > j 2 N (either of which could also be zero) so that ai = aj, which means that ai−j = e. Since i > j, we have i − j > 0. If i − j = 1, then ai = ai+1 and this implies a = e, so a−1 = a = e 2 H. Otherwise, if i − j ≥ 2 then i − j − 1 ≥ 1, and so ai−j−1 = a−1 2 H. 4. (a) Let H ≤ Z. Then we know that H is cyclic by the fundamental theorem of cyclic groups, as Z =< 1 > is cyclic. Let t be the smallest positive integer in H. Such a number exists because H is assumed to be nontrivial, i.e., there exists a non-identity element in H. Since t 2 H, and H is finite, kt = 0 for some nonzero integer k. But this implies that t = 0, contradicting the fact that t is a positive integer. (b) Let x 2 Z(G). Then we note that xa = ax since xy = yx for any y in G, namely, y = a. Thus Z(G) ⊆ C(a) for any a 2 G. Then we use the 2 step subgroup test. Notice that since Z(G) ≤ G, if y 2 Z(G) then y−1 2 Z(G). Thus for x; y 2 Z(G), and any element z 2 G, we have xyz = xzy = zxy, so that xy 2 Z(G). Thus Z(G) ≤ C(a) for any a. (We remark that this proof doesn't seem to use the fact that our ambient group is C(a), except that we checked that Z(G) ⊆ C(a) at the start.) See number 3(b)(ii) for an example where Z(G) 6= C(a) for some a, or, refer to your class notes where we computed the centralizer C(F ) in D4, and Z(D4) is a proper subgroup of C(F ). (c) (i) The number of subgroups of the cyclic group Z32 is the number of positive divisors of 32, which is 6: 1; 2; 22; 23; 24; 25. (ii) Each of the subgroups are cyclic because any subgroup of a cyclic group is cyclic (this is one of the statements of the Fundamental Theorem of Cyclic Groups). (iii) Let k be a positive divisor of 32. Then each cyclic group of order k is generated by the element 32=k. 2 −1 (d) (i) Suppose σ = τ1 ··· τr, where the τi are transpositions. Notice that τi = τi, so that (by repeated applications of the so-called \socks and shoes principle"), −1 −1 −1 σ = τr ··· τ1 = τr ··· τ1, the product of the same number of transpositions. The result follows. (ii) Suppose β is odd, then so is β−1 (by the above). Then if α is even, then βαβ−1 is the product of two odd and one even permutation, which is even. If α is odd, then βαβ−1 is the product of three odd permutations, which is odd. Now, if β is even, we apply the same argument. If α is even, then βαβ−1 is the product of three even permutations, which is even. If α is odd, then βαβ−1 is the product of one odd and two even permutations, which is also odd. In every case, α and βαβ−1 have the same parity. 3.
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