Show that if \(H \leq N_G(K)\) then \(HK \leq G\), and give a counterexample showing that this condition is necessary.
Solution
Let \(H,K\le G\) and put \(HK=\{hk:h\in H,\ k\in K\}\).
1. If \(H\le N_G(K)\), then \(HK\) is closed under multiplication.
Let \(h_1k_1,h_2k_2\in HK\). Then \[(h_1k_1)(h_2k_2)=(h_1h_2)\bigl(h_2^{-1}k_1h_2\bigr)k_2 .\] Since \(h_2\in N_G(K)\), \(h_2^{-1}k_1h_2\in K\), so the product lies in \(HK\).
2. If \(H\le N_G(K)\), then \(HK\) is closed under inverses.
For \(hk\in HK\), \((hk)^{-1}=k^{-1}h^{-1}=h^{-1}\bigl(hk^{-1}h^{-1}\bigr)\), and \(hk^{-1}h^{-1}\in K\) because \(h\in N_G(K)\).
3. If \(H\le N_G(K)\), then \(HK\le G\).
\(1=1\cdot1\in HK\), and steps 1 and 2 give the subgroup criterion.
4. In \(G=S_3\), the subgroups \(H=\langle(1\,2)\rangle\) and \(K=\langle(1\,3)\rangle\) satisfy \(H\not\le N_G(K)\), and \(HK\) is not a subgroup.
\((1\,2)(1\,3)(1\,2)=(2\,3)\notin K\), so \((1\,2)\notin N_G(K)\). Since \(H\cap K=1\), \(|HK|=|H||K|/|H\cap K|=4\), and \(4\nmid 6=|S_3|\), so by Lagrange’s theorem \(HK\) is not a subgroup.
5. Q.E.D.
Step 3 proves the implication, and step 4 shows that it fails without the hypothesis \(H\le N_G(K)\).