2.12 Order of product of subgroups
In this section \(G\) is a group with identity element \(e\), and \(H\) and \(K\) are two subgroups of \(G\).
Suppose \(H\) and \(K\) are finite. Then
In some important cases, \(HK\leq G\). But often it is not. (We’ll see examples.)
Properly interpretted, Theorem 2.93 is also true when \(H\) and \(K\) are infinite.
Let \(H\times K\) denote the Cartesian product of \(H\) and \(K\). For each \(h\in H\) and \(k\in K\), \(hk\in HK\). So we can define a function \(f:H\times K\to HK\) by setting \(f(h,k) = hk\). Clearly, \(f\) is onto. So, by Proposition 2.77, \(P(f) = \{ f^{-}(g) : g\in HK\} \) is a partition of \(H\times K\).
I claim that, for each \(g\in HK\), \(|f^{-1}(g)| = |H\cap K|\). To see this, pick \(h\in H\) and \(k\in K\) such that \(g = hk\). Then, for \(u\in H\cap K\), \(hu\in H\), \(u^{-1}k\in K\) and \(f(hu,u^{-1}k) = hk = g\). So we can define a map \(\phi :H\cap K\to f^{-1}(g)\) by setting \(\phi (u) = (hu,u^{-1}k)\).
I claim that \(\phi \) is both one-one and onto. To see that \(\phi \) is one-one, suppose \(u_1,u_2\in H\cap K\) and \(\phi (u_1) = \phi (u_2)\). Then \(u_1 = h^{-1}hu_1 = h^{-1}hu_2 = u_2\).
To see that \(\phi \) is onto, suppose \((h_1,k_1)\in f^{-1}(g)\). Then \(h_1\in H\), \(k_1\in K\) and \(h_1k_1 = g = hk\). It follows that \(h^{-1}h_1 = kk_1^{-1}\). But then \(u :=h^{-1}h_1 = kk_1^{-1}\in H\cap K\). And \(\phi (u) = (hu, u^{-1}k) = (hh^{-1}h_1, k_1k^{-1}k) = (h_1,k_1)\). As \((h_1,k_1)\) was an arbitrary element of \(f^{-1}(g)\), this proves that \(\phi :H\cap K\to f^{-1}(g)\) is onto.
Since \(\phi \) is one-one and onto, \(|f^{-1}(g)| = |H\cap K|\) for every \(g\in HK\). So \(P(f)\) partitions \(H\times K\) into \(|HK|\) sets each of size \(|H\cap K|\). It follows that \(|H||K| = |H\times K| = |HK||H\cap K|\). So Theorem 2.93 is proved.
Let \(G = S_3\) and let \(H = \langle \sigma \rangle \), \(K = \langle \tau \rangle \) where
Then \(|H| = |\sigma | = 2 = |\tau | = |K|\). So \(H = \{ e, \sigma \} \), \(K = \{ e, \tau \} \) and \(H\cap K = \{ e\} \). Therefore, by Theorem 2.93, we see that \(|HK| = 4\). As \(4 \nmid 6 = |G|\), Lagrange’s theorem shows that \(HK\) is not a subgroup of \(G\). And, in fact, since \(|HK| {\gt} |G|/2\), it follows that the smallest subgroup of \(G\) containting \(HK\) is \(G\) itself. Therefore \(G = \langle \sigma , \tau \rangle \).