- Show that if \(H \leq N_G(K)\) then \(HK \leq G\), and give a counterexample showing that this condition is not necessary.
Solution
1. If \(H\le N_G(K)\), then \(HK\) is a subgroup of \(G\).
For \(h\in H\) and \(k\in K\), the normalizer hypothesis gives \[ hkh^{-1}\in K. \] Hence \[ hk=(hkh^{-1})h\in KH, \] so \(HK\subseteq KH\). Replacing \(h\) by \(h^{-1}\) gives the reverse inclusion, hence \[ HK=KH. \] For subgroups \(H,K\le G\), the set \(HK\) is a subgroup exactly when \(HK=KH\) (P-76B6F). Explicitly, if \[ h_1k_1,h_2k_2\in HK, \] then \(k_1h_2\in KH=HK\), so their product lies in \(HK\); similarly inverses lie in \(HK\). Therefore \(HK\le G\).
2. The normalizer condition is not necessary.
Take \[ G=S_3, \qquad H=\langle(123)\rangle, \qquad K=\langle(12)\rangle. \] The subgroups have orders \(3\) and \(2\) and intersect trivially, so \[ |HK|=\frac{|H||K|}{|H\cap K|}=6=|S_3|. \] Thus \[ HK=S_3, \] which is a subgroup.
However, \[ (123)(12)(123)^{-1}=(23)\notin K, \] so \((123)\notin N_G(K)\) and hence \[ H\not\le N_G(K). \] Thus \(H\le N_G(K)\) is sufficient but not necessary for \(HK\) to be a subgroup.