Let $G$ be a group which is the product $G=NH$ of subgroup $N,H\subset G$ where $N$ is normal.
Let $N\cap H=\{1\}$.
I am trying to show that that there is an iso $G\cong N\rtimes H$, with the automorphism by conjugation on itself.
What isomorphism could I use? I tried various different ones but most weren't even homomorphisms.
On
Let the elements of $N \rtimes H$ be expressed as ordered pairs $(n,h)$, subject to the usual multiplication rule $(n,h)(n',h') = (nhn'h^{-1},hh')$.
Define a map $f : N \rtimes H \to G$ by the formula $f(n,h)=nh$.
Check that $f$ is a homomorphism (should be easy).
$f$ is onto since $G=NH$.
$f$ is one-to-one since $N \cap H = \{1\}$.
So $f$ is an isomorphism.
The homomorphism
$$\phi:H\to\text{Aut}\,N\;,\;\;\phi(h)(n):=n^h:=hnh^{-1}\;\;\;\text{does the trick}$$