Semidirect Product of Groups is Group

Theorem
Let $H$ and $N$ be groups.

Let $\operatorname{Aut}(N)$ denote the automorphism group of $N$.

Let $\phi : H\to \operatorname{Aut}(N)$ be a group homomorphism, that is, let $H$ act on $N$.

Let $N\rtimes_\phi H$ be the semidirect product of $N$ and $H$ with respect to $\phi$, that is:


 * $N\rtimes_\phi H = (N\times H, \circ)$ where
 * $(n_1, h_1) \circ (n_2, h_2) = (n_1\cdot \phi(h_1)(n_2), h_1\cdot h_2)$

Then $N\rtimes_\phi H$ is a group.

Associativity
Let $(n_1,h_1),(n_2,h_2),(n_3,h_3)\in N\times H$. Then

Identity Element
Let $e = (e_N,e_H)$ and $(n,h)\in N\times H$.

Then

and

Thus $e$ is an identity element for $\circ$.

Existence of inverse element
Let $(n,h)\in N\times H$.

We are looking for $(m,g)$ with $(n,h)\circ(m,g) = (e_N,e_H) = (m,g)\circ(n,h)$, that is:
 * $\begin{eqnarray}

n \phi_h(m) &= e_N \\ hg &= e_H \\ m \phi_g(n) &= e_N \\ gh &= e_H \end{eqnarray}$

Thus we take $g=h^{-1}$.

From the first equation we have $m=\phi_h^{-1}(n^{-1})$; from the third $m=\phi_{h^{-1}}(n)^{-1}$.

These are equal because $H$ acts by automorphisms.

The inverse of $(n,h)$ is thus $(\phi_{h^{-1}}(n^{-1}),h^{-1})$.