- Let \(H\leq G\) be a subgroup and \(N\normal G\) be a normal subgroup. Show that \(NH \leq G\) is a subgroup.
Solution
We use the one-step subgroup criterion.
1. The set \(NH\) is nonempty.
Since \(e\in N\cap H\), one has \(e=ee\in NH\).
2. If \(n_1h_1,n_2h_2\in NH\), then\[ (n_1h_1)(n_2h_2)^{-1}\in NH. \]
Compute \[ (n_1h_1)(n_2h_2)^{-1} =n_1h_1h_2^{-1}n_2^{-1}. \] Put \(h=h_1h_2^{-1}\in H\). Since \(N\trianglelefteq G\), \[ hn_2^{-1}h^{-1}\in N. \] Hence \[ n_1h n_2^{-1} =n_1(hn_2^{-1}h^{-1})h, \] which is a product of an element of \(N\) and an element of \(H\), so it lies in \(NH\).
3. Therefore \(NH\le G\).
Apply the subgroup criterion using 1 and 2.