Semidirect Product of Groups is Group

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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.


Proof

Associativity

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

\(\displaystyle ((n_1, h_1) \circ (n_2, h_2)) \circ (n_3, h_3)\) \(=\) \(\displaystyle ((n_1\cdot \phi_{h_1}(n_2), h_1\cdot h_2)) \circ (n_3,h_3)\)
\(\displaystyle \) \(=\) \(\displaystyle (n_1\cdot \phi_{h_1}(n_2) \cdot \phi_{h_1h_2}(n_3), h_1\cdot h_2\cdot h_3)\)
\(\displaystyle \) \(=\) \(\displaystyle (n_1\cdot \phi_{h_1}(n_2) \cdot \phi_{h_1}(\phi_{h_2}(n_3)), h_1\cdot h_2\cdot h_3)\) Definition of Group Action
\(\displaystyle \) \(=\) \(\displaystyle (n_1\cdot \phi_{h_1}(n_2\phi_{h_2}(n_3)), h_1\cdot h_2\cdot h_3)\) $H$ acts by automorphisms
\(\displaystyle \) \(=\) \(\displaystyle (n_1, h_1) \circ ((n_2, h_2) \circ (n_3, h_3))\)

$\Box$

Identity Element

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

Then

\(\displaystyle e \circ (n,h)\) \(=\) \(\displaystyle (e_N\cdot \phi_{e_H}(n), e_H\cdot h)\)
\(\displaystyle \) \(=\) \(\displaystyle (n, h)\) Definition of Group Action

and

\(\displaystyle (n,h) \circ e\) \(=\) \(\displaystyle (n\cdot \phi_{h}(e_N), h\cdot e_H)\)
\(\displaystyle \) \(=\) \(\displaystyle (n, h)\) $\phi_h(e_N)=e_N$ because $H$ acts by automorphisms

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

$\Box$

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 $(\phi_{h^{-1}}(n^{-1}),h^{-1})$.

$\Box$

$\blacksquare$