Inverse of Element in Semidirect Product

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Theorem

Let $N$ and $H$ be groups.

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

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

Let $(n,h)\in N\rtimes_\phi H$.


Then

$\begin{align*}(n,h)^{-1} &= (\phi_{h^{-1}}(n^{-1}), h^{-1}) \\ &= (\phi_{h^{-1}}(n)^{-1}, h^{-1}) \\ &= (\phi_{h}^{-1}(n^{-1}), h^{-1}) \\ &= (\phi_{h}^{-1}(n)^{-1}, h^{-1}) \end{align*}$


Proof

Follows from Semidirect Product of Groups is Group.

The alternatives follow from the fact that $H$ acts by automorphisms.

$\blacksquare$