MATH 351 – Modern Algebra 1 Selected Solutions for Problem Set 7 November 3, 2018

4.46. (1) We are given that N and K are subgroups of index 2 of G with N ≠ K. By Theorem 4.1 in Lee, note that both N an K are normal subgroups of G. By the result of Exercise 4.5, N ∩ K is also a normal subgroup of G. This follows because if g ∈ G is arbitrary, and a ∈ N ∩ K, then a ∈ N, so gag^{−1} ∈ N and similarly a ∈ K, so gag^{−1} ∈ K. Hence gag^{−1} ∈ N ∩ K, so N ∩ K is normal in G. The same argument just given also works if we only take g ∈ N. Hence N ∩ K is also a normal subgroup of N. Now by the Second Isomorphism Theorem, we have

N/(N ∩ K) ≅ NK/K.

We claim that NK = G. This is true because K has index 2 in G. In the set

NK = {nk | n ∈ N, k ∈ K},

if n ∈ K as well, then nK = K, so we get K ⊂ NK. On the other hand, we are assuming N ≠ K, so there are n ∈ N that are in G\K as well. If n ∉ K, then the coset nK = G\K, the other left coset of K in G. So G\K ⊂ NK as well. Hence K ∪(G\K) = G ⊆ NK ⊆ G. It follows that NK = G. Now we have proved what was required since

[N : N ∩ K] = |N/(N ∩ K)| = |NK/K| = |G/K| = 2.

(2) Now consider the mapping

\begin{matrix}α : G & → G/N × G/K \\ g & ↦ (gN, gK) \end{matrix}

G/N and G/K have order 2 so each is isomorphic to Z_{2} by Corollary 4.2. We want to show that α is a surjective group homomorphism and identify the kernel as N ∩ K to show that G/(N ∩ K) ≅ Z_{2} × Z_{2}. Here are the individual steps:

• We claim first that α is a group homomorphism. This is true because

α(g·h) = ((g·h)N, (g·h)K) = (gN ·hN, gK ·hK) = (gN, gK)·(hN, hK) = α(g)·α(h),

using the definition of the coset product in G/N and G/K.

• The kernel of α consists of the g ∈ G such that (gN, gK) = (N, K), the identity element of G/N × G/K. But (gN, gK) = (N, K) if and only if g ∈ N and g ∈ K, so g ∈ N ∩ K. Hence ker(α) = N ∩ K.

• Next, we claim that α(G) = (G/N, G/K) (that is, that α is surjective). This follows by thinking about what part (1) of the problem says. We have [N : N ∩ K] = 2. The same argument, but reversing the roles of N, K shows that [K : N ∩ K] = 2 as well.

Moreover [G : N] = [G : K] = 2 by assumption. It follows that G decomposes into four disjoint subsets:

N ∩ K, (G \ N) ∩ K, N ∩ (G \ K), and (G \ N) ∩ (G \ K).

All of these are nonempty since [N : N ∩ K] = [K : N ∩ K] = 2. The g in the first map to (N, K) under α, and the g in the other three subsets map to the other three elements in G/N × G/K. Therefore α is surjective.

Comment: We actually saw an example of the pattern described here when we considered G = D_{8}, N = ⟨R_{90}⟩ and K = {R_{0}, R_{180}, F_{1}, F_{2}}. These are two unequal subgroups of order 4 in a group of order 8, so we’re exactly in the situation of the problem. In this case, N ∩ K = {R_{0}, R_{180}} = Z(D_{8}) and we showed directly that

D_{8}/Z(D_{8}) ≅ Z_{2} × Z_{2}.