Inverse of Element in Semidirect Product

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 $\tuple {n, h} \in N \rtimes_\phi H$.

Then:

\(\ds \tuple {n, h}^{-1}\)

\(=\)

\(\ds \tuple {\map {\phi_{h^{-1} } } {n^{-1} }, h^{-1} }\)

\(\ds \)

\(=\)

\(\ds \tuple {\paren {\map {\phi_{h^{-1} } } n}^{-1}, h^{-1} }\)

\(\ds \)

\(=\)

\(\ds \tuple {\map { {\phi_h}^{-1} } {n^{-1} }, h^{-1} }\)

\(\ds \)

\(=\)

\(\ds \tuple {\paren {\map { {\phi_h}^{-1} } n}^{-1}, h^{-1} }\)

Proof

Follows from Semidirect Product of Groups is Group.

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

$\blacksquare$

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