Semidirect Product is Abelian iff Components are Abelian and Action is Trivial

Theorem

Let $N$ and $H$ be groups.

Let $H$ act by automorphisms on $N$ via $\phi$.

Let $N \rtimes_\phi H$ be the corresponding semidirect product.

Then the following are equivalent:

$(1): \quad N \rtimes_\phi H$ is abelian

$(2): \quad N$ and $H$ are abelian and $H$ acts trivially

Proof

$(1)$ implies $(2)$

Let $n \in N$, $h \in H$.

From $\tuple {n, e} \tuple {e, h} = \tuple {e,h} \tuple {n, e}$ we have $n \map {\phi_e} e = e \map {\phi_h} n$.

Thus $H$ acts trivially.

By Semidirect Product with Trivial Action is Direct Product, $N \rtimes_\phi H = N \times H$.

By External Direct Product of Abelian Groups is Abelian Group, $N$ and $H$ are abelian.

$\Box$

$(2)$ implies $(1)$

By Semidirect Product with Trivial Action is Direct Product, $N \rtimes_\phi H = N \times H$.

By External Direct Product of Abelian Groups is Abelian Group, $N \rtimes_\phi H$ is abelian.

$\blacksquare$