## Outer Semidirect Product

Given any two groups $N$ and $H$ and a group homomorphism $\phi:H\to\text{Aut}(N)$, we can construct a new group $N\rtimes_\phi H$, called the (outer) semidirect product of $N$ and $H$ with respect to $\phi$, defined as follows.
(i) The underlying set is the Cartesian product $N\times H$.
(ii) The operation, $\bullet$, is determined by the homomorphism $\phi$:

$\bullet: (N\rtimes_\phi H)\times (N\rtimes_\phi H)\to N\rtimes_\phi H$

$(n_1,h_1)\cdot(n_2,h_2)=(n_1\phi_{h_1}(n_2),h_1h_2)$

for $n_1,n_2\in N$ and $h_1,h_2\in H$.

This defines a group in which the identity element is $(e_N, e_H)$ and the inverse of the element $(n,h)$ is $(\phi_{h^{-1}}(n^{-1}), h^{-1})$.
Pairs $(n, e_H)$ form a normal subgroup isomorphic to $N$, while pairs $(e_N, h)$ form a subgroup isomorphic to $H$.

## Inner Semidirect Product (Definition)

Given a group $G$ with identity element $e$, a subgroup $H$, and a normal subgroup $N\lhd G$; then the following statements are equivalent:

(i) $G$ is the product of subgroups, $G=NH$, where the subgroups have trivial intersection, $N\cap H=\{e\}$.
(ii) For every $g\in G$, there are unique $n\in N$ and $h\in H$, such that $g=nh$.

If these statements hold, we define $G$ to be the semidirect product of $N$ and $H$, written $G=N\rtimes H$.

## Inner Semidirect Product Implies Outer Semidirect Product

Suppose we have a group $G$ with $N\lhd G$, $H\leq G$ and every element $g\in G$ can be written uniquely as $g=nh$ where $n\in N$, $h\in H$.

Define $\phi: H\to\text{Aut}(N)$ as the homomorphism given by $\phi(h)=\phi_h$, where $\phi_h(n)=hnh^{-1}$ for all $n\in N$, $h\in H$.

Then $G$ is isomorphic to the semidirect product $N\rtimes_{\phi}H$, and applying the isomorphism to the product, $nh$, gives the tuple, $(n,h)$. In $G$, we have
$\displaystyle (n_1h_1)(n_2h_2)=n_1h_1n_2(h_1^{-1}h_1)h_2=(n_1\phi_{h_1}(n_2))(h_1h_2)=(n_1,h_1)\cdot(n_2,h_2)$
which shows that the above map is indeed an isomorphism.