The subgroup \(H^x K^y\), where \(H,K \le G\) and \(x,y \in G\) (FGT 1A.4)

Suppose that \(G = HK\), where \(H\) and \(K\) are subgroups. Show that also \(G = H^x K^y\) for all elements \(x,y \in G\). Deduce that if \(G = HH^x\) for a subgroup \(H\) and an element \(x \in G\), then \(H = G\).

We first show that \(HK = HxK\). Firstly we can write \(x = h_0 k_0\) for \(h_0 \in H\) and \(k_0 \in H\). Clearly for any \(hk \in HK\) there is some \(h_1 \in H\) and \(k_1 \in K\) such that \(hk = h_1 h_0 k_0 k_1  = h_1 x k_1 \in HxK\), and the opposite is true as well. Furthermore we know that \(H^x\) is a subgroup with order \(\left|H\right|\), so \(\left|H^xK\right| = \frac{\left|H^x\right|\left|K\right|}{\left|H^x \cap K\right|} = \frac{\left|H\right|\left|K\right|}{\left|H^x \cap K\right|} = \left|HxK\right| = \left|G\right|\) and so \(G = H^xK\). We can do likewise with \(K^y\) to obtain the conclusion.

If \(G = HH^x\), then \(G = H(H^x)^{x^{-1}} = H\).