## § Internal versus External semidirect products

Say we have an inner semidirect product. This means we have subgroups $N, K$ such that $NK = G$, $N$ normal in $G$ and $N \cap K = \{ e \}$. Given such conditions, we can realize $N$ and $K$ as a semidirect product, where the action of $K$ on $N$ is given by conjugation in $G$. So, concretely, let's think of $N$ (as an abstract group) and $K$ (as an abstract group) with $K$ acting on $N$ (by conjugation inside $G$). We write the action of $k$ on $n$ as $n^k \equiv knk^{-1}$. We then have a homomorphism $\phi: N \ltimes K \rightarrow G$ given by $\phi((n, k)) = nk$. To check this is well-defined, let's take $s, s' \in N \ltimes K$, with $s \equiv (n, k)$ and $s' \equiv (n', k')$. Then we get:
\begin{aligned} &\phi(ss') = \\ &=\phi((n, k) \cdot (n', k')) \\ &\text{definition of semidirect product via conjugation:} \\ &= \phi((n {n'}^k, kk')) \\ &\text{definition of \phi:} \\ &= n n'^{k} kk' \\ &\text{definition of n'^k = k n' k^{-1}:} \\ &= n k n'k^{-1} k k' \\ &= n k n' k' \\ &= \phi(s) \phi(s') \end{aligned}
So, $\phi$ really is a homomorphism from the external description (given in terms of the conjugation) and the internal description (given in terms of the multiplication). We can also go the other direction, to start from the internal definition and get to the conjugation. Let $g \equiv nk$ and $g' \equiv n'k'$. We want to multiply them, and show that the multiplication gives us some other term of the form $NK$:
\begin{aligned} gg' \\ &= (n k) (n' k') \\ &= n k n' k' \\ &= \text{insert k^{-1}k: } \\ &= n k n' k^{-1} k k' \\ &= n (k n' k^{-1}) k k' \\ &= \text{N is normal, so k n' k^{-1} is some other element n'' \in N:} \\ &= n n'' k k' \\ &= N K \end{aligned}
So, the collection of elements of the form $NK$ in $G$ is closed. We can check that the other properties hold as well.