Let \(H,K\le G\) and suppose \(H\le N_G(K)\). Prove that \(HK\) is a subgroup of \(G\). Give a counterexample if \(K\) is only a subset, not a subgroup.
Solution
1. If \(H\le N_G(K)\), then \(HK\le G\).
The identity belongs to \(HK\). Let \(h_1k_1,h_2k_2\in HK\). Since \(h_2\in N_G(K)\), \[ k_1h_2=h_2(h_2^{-1}k_1h_2) \] with \(h_2^{-1}k_1h_2\in K\). Hence \[ (h_1k_1)(h_2k_2)=h_1h_2\,(h_2^{-1}k_1h_2)k_2\in HK. \] Also, for \(hk\in HK\), \[ (hk)^{-1}=k^{-1}h^{-1}=h^{-1}(hk^{-1}h^{-1})\in HK \] because \(h^{-1}\in N_G(K)\). Therefore \(HK\) is a subgroup.
2. The subgroup hypothesis on \(K\) is necessary.
Take the additive group \(G=\ZZ\), let \(H=2\ZZ\), and let \(K=\{0,1\}\). Since \(G\) is abelian, every element normalizes every subset in the setwise sense, but \[ H+K=2\ZZ\cup(2\ZZ+1)=\ZZ. \] This particular choice still gives a subgroup, so choose instead \(H=\{0\}\) and \(K=\{0,1\}\). Then \(H+K=\{0,1\}\) is not a subgroup of \(\ZZ\) because \(1+1=2\notin\{0,1\}\).