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.