Semidirect product

Remarks

The identity is $(e, e)$, and the inverse of $(n, h)$ is $(\varphi_{h^{-1}}(n^{-1}), h^{-1})$. The set $\{(n, e) : n \in N\}$ is a normal subgroup isomorphic to $N$, and $\{(e, h) : h \in H\}$ is a subgroup isomorphic to $H$. If $\varphi$ is the trivial homomorphism, then $N \rtimes_\varphi H$ is the direct product $N \times H$. The dihedral group $D_n$ is $\mathbb{Z}/n\mathbb{Z} \rtimes_\varphi \mathbb{Z}/2\mathbb{Z}$, where the nonzero element of $\mathbb{Z}/2\mathbb{Z}$ acts by $x \mapsto -x$.