Homework 2 due 2/07/02 2.1 Remak-Krull-Schmidt. (4.47) If K E G then show that G satisfies both chain conditions (for normal subgroups) if both K and G=K satisfy both chain conditions. We can assume that G is infinite since the chain conditions are obviously both satisfies for finite groups. Ans: Let Á : G ! G=K be the quotient map and suppose that fGig = fG1 ¸ G2 ¸ G3 ¢ ¢ ¢ g is a decreasing sequence of normal subgroups of G . Then fÁ(Gi)g is a decreasing sequence of normal subgroups of G=K and fGi\Kg is a decreasing sequence of normal subgroups of K. So both of these become stationary for large enough i, say i ¸ n. By the following lemma this implies that Gn = Gn+1 = Gn+2 = ¢ ¢ ¢ . This proves the DCC. The ACC is similar. Lemma 2.1 (Kyle’s Lemma). If A ¸ B, Á(A) = Á(B) and A\K = B\K then A = B. Proof. It suffices to show that A · B. So take any a 2 A. Then Á(a) 2 Á(A) = Á(B) so there is a b 2 B · A such that Á(a) = Á(b). This implies that Á(ab¡1) = 1, i.e., ab¡1 2 ker Á = K. But ab¡1 is also an element of A since a; b both lie in A. Thus ab¡1 2 K \ A = K \ B so ab¡1 2 B ) a 2 Bb = B. Partially conversely, if G satisfies both chain conditions then show that any quotient also satisfies both chain conditions. This is obvious since sequences of normal subgroups of G=K correspond to sequences of normals subgroups of G which contain K. (4.48) If G satisfies both chain conditions and G £ G »= H £ H then show that G »= H. [Note: use RKS but notice that you are not assuming that H satisfies the chain conditions.] Ans: First we need to show that H satisfies both chain conditions. This follows from (4.47) above. Since G satisfies both chain conditions, so does G £ G since G is a normal subgroup of G £ G with quotient G. Since H is a quotient of H £ H »= G £ G, H also satisfies both chain conditions by the second part of (4.47). By the same argument H £ H also satisfies both chain conditions. The RKS theorem now applies to H and H £ H. So both G and H are products of indecomposable groups. In G£G »= H£H, these indecomposable groups occur twice. By RKS, they occur the same number of times in each decomposition. Dividing these numbers by 2 we conclude that G »= H. 1 2.2 p-Groups Exercise 1: If P is a nonabelian group of order p3 then show that P=Z(P ) is elementary abelian of order p2. Ans: We know that P=Z(P ) is not cyclic so it’s order must be at least p2. But Z(P ) is nontrivial so it must have order p and index p2. The quotient group P=Z(P ) has order p2 so we know that it is abelian. Since it is not cyclic it must be Z=p £ Z=p. Exercise 2: Let Φ(P ) be the intersection of all maximal subgroups of a p- group P . [Φ(P ) is called the Frattini subgroup P .] Show that P=Φ(P ) is elementary abelian. Ans: The Frattini subgroup Φ(P ) is the kernel of Y Y » P ¡! H = G=Mi = Z=p Mi<G max Q Thus P=Φ(P ) is a subgroup of H »= Z=p. But any subgroup of an elemen- tary abelian group is obviously elementary abelian. [Anton’s proof] For any maximal subgroup M of G, G=M »= Z=p. So: 1. M contains G0. 2. M contains gp for any g 2 G. Since this is true for each maximal M we have: T 1. Φ(P ) = M contains G0. 2. Φ(P ) contains gp for any g 2 G. The first statement implies that P=Φ(P ) is abelian. The second statement implies that every element has order p so it is elementary abelian. 2.3 Applications of Sylow Ex. 3: Suppose that G is a group of order p2q where q ´= 1 mod p and p2 ´= 1 mod q. Then show that G is abelian. Ans: The index of N(Q) (where Q is a q-Sylow subgroup of G) cannot be p or p2 since neither of these numbers is congruent to 1 modulo q. Con- sequently, the index is 1 and Q C G. Similarly, q ´= 1 mod p implies P C G where P is the p-Sylow subgroup of G. But P \ Q = 1 so G = P £ Q. Since both P and Q are abelian (groups of order p2 are abelian), G = P £ Q is abelian. Ex. 4: Find all subgroups of D12 (the dihedral group of order 12). 6 2 ¡1 Ans: The dihedral group D12 = ha; bja = 1 = b ; bab = a i has a cyclic normal subgroup C = hai = f1; a; ¢ ¢ ¢ ; a5g of order 6. 2 1. C has four subgroups 1; ha3i ; ha2i ;C of order 1; 2; 3; 6. Any other sub- group H of D12 must intersect C in one of these subgroups. i 2. Every element ba of D12 not in C has order 2 and generates a subgroup i Hi = f1; ba g. These are the subgroups satisfying Hi \ C = 1 3 3 3. If H \ C = ha i (and HC 6= D12) then H = Hi ha i. There are three of these subgroups: 3 3 3 3 (a) H0 ha i = H3 ha i = f1; a ; ba ; bg 3 3 3 4 (b) H1 ha i = H4 ha i = f1; a ; ba; ba g 3 3 3 2 5 (c) H2 ha i = H5 ha i = f1; a ; ba ; ba g 2 2 4. If H \ C = ha i (and HC 6= D12) then H = Hi ha i. There are two of these subgroups: 2 2 2 2 4 2 4 (a) H0 ha i = H2 ha i = H4 ha i = f1; a ; a ; b; ba ; ba g 2 2 2 2 4 3 5 (b) H1 ha i = H3 ha i = H5 ha i = f1; a ; a ; ba; ba ; ba g 5. Finally, if H \ D12 = C (and H 6= C) then H = D12. This makes a total of 4 + 6 + 3 + 2 + 1 = 16 subgroups of D12. With a similar analysis we see that, in general, D2n has X (d + 1) djn ­ ® ­ ® subgroups and they have the form ad where djn and ad; bai where djn and 0 · i < d. 3.