BRIEF ANSWERS FOR TEST 2

1. Suppose that F is an automorphism of G/H. Then F maps the identity coset H to some coset aH, a\in G. But this information (the element a) already determines F completely, since for any other coset we must have F(gH) = gF(H) = gaH. Denote the F that is determined in this way by the element a\in G by F_a. In order for F_a to be well defined (as the function F_a(gH) = gaH) we must have that gH = hH implies gaH = haH. Equivalently, we must have that h^{-1}g in H implies that a^{-1}h^{-1}ga in H. Equivalently we must have that h^{-1}g in H implies that h^{-1}g in aHa^{-1}. Equivalently we must have that H is contained in aHa^{-1}. Thus F_a is well defined iff a is in N_G(H). Because of these remarks, and F_a(F_b(x)) = F_{ba}(x), the function N_G(H) --> Aut(G/H) : a mapsto F_{a^{-1}} is a surjective group homomorphism. The kernel is easily seen to be H, so N_G(H)/H is isomorphic to Aut(H).

2. There are two abelian groups of order 18 (Z_18 and Z_3 x Z_6), and three nonabelian groups of order 18 (D_9, Z_3 x D_3, and a semidirect product of Z_3 x Z_3 by Z_2 determined by the homomorphism from Z_2 to Aut(Z_3 x Z_3) = GL_2(Z_3) that maps the nonidentity element of Z_2 to the negative of the identity matrix). This can be proved by techniques used in class.

3. G has 4 Sylow 3-subgroups, and the action by conjugation on these subgroups induces a homomorphism rho from G into S_4. Since no Sylow 3-subgroup normalizes any other, each element g of G that has order three must map to a 3-cycle. Indeed, the eight elements of G of order three must map to the eight elements of order three in S_4, so the image of rho contains the subgroup of S_4 generated by the 3-cycles. This subgroup is A_4. Thus, the image of rho can only be S_4 or A_4.

   The conclusions up to this point were based only on the assumption that G has no normal Sylow 3 subgroup. To finish we must use the assumption that G has no normal Sylow 2 subgroup (Since A_4 x Z_2 has order 24, has no normal Sylow 3-subgroup, but is not isomorphic to S_4.)

   If the image of rho is A_4, then rho has a 2-element kernel N, and G/N is isomorphic to A_4. Since the Klein group V is a normal subgroup of order 4 in A_4, the subgroup rho^{-1}(V) is a normal subgroup of G of order 8. It would be a normal Sylow 2-subgroup of G, and by assumption there is no such subgroup. Thus rho does not have image A_4, and must therefore be an isomorphism between G and S_4.

4. 315 = 3^2*5*7. Let T be a Sylow 3-subgroup, F be a Sylow 5-subgroup, and S be a Sylow 7-subgroup. Since T is assumed to be normal, T join F is a subgroup of order 3^2*5, and this join is a semidirect product of T by F. Since |F|=5 does not divide |Aut(Z_9)|=6 or |Aut(Z_3 x Z_3)| = 48, the semidirect product is direct. Hence elements of F commute with elements of T. The same argument shows that elements of S commute with elements of T. Of course, elements of T commute with T, since T has order 9. Thus, the centralizer of T contains T, F and S, hence T is in the center of G. G/T has order 35, and every group of order 35 is cyclic. Thus, since the quotient modulo a central subgroup is cyclic, G is abelian.