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

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 φ : G → G/K be the quotient map and suppose that

{Gi} = {G1 ≥ G2 ≥ G3 ···} is a decreasing sequence of normal subgroups of G . Then {φ(Gi)} is a decreasing sequence of normal subgroups of G/K and {Gi∩K} 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, φ(A) = φ(B) and A∩K = B∩K then A = B.

Proof. It suﬃces to show that A ≤ B. So take any a ∈ A. Then φ(a) ∈ φ(A) = φ(B) so there is a b ∈ B ≤ A such that φ(a) = φ(b). This implies that φ(ab−1) = 1, i.e., ab−1 ∈ ker φ = K. But ab−1 is also an element of A since a, b both lie in A. Thus ab−1 ∈ K ∩ A = K ∩ B so ab−1 ∈ B ⇒ a ∈ 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

Mimaximal subgroup M of G, G/M ∼= Z/p. So: 1. M contains G0. 2. M contains gp for any g ∈ 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 ∈ 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, b|a = 1 = b , bab = a i has a cyclic normal subgroup C = hai = {1, a, ··· , a5} of order 6.

2 1. C has four subgroups 1, ha3i , ha2i ,C of order 1, 2, 3, 6. Any other subgroup 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 = {1, ba }. 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 = {1, a , ba , b} 3 3 3 4 (b) H1 ha i = H4 ha i = {1, a , ba, ba } 3 3 3 2 5 (c) H2 ha i = H5 ha i = {1, a , ba , ba }

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 = {1, a , a , b, ba , ba } 2 2 2 2 4 3 5 (b) H1 ha i = H3 ha i = H5 ha i = {1, a , a , ba, ba , ba }

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) d|n ® ® subgroups and they have the form ad where d|n and ad, bai where d|n and 0 ≤ i < d.