# Inverse of Element in Semidirect Product

Jump to navigation
Jump to search

## 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.

$\blacksquare$

This needs considerable tedious hard slog to complete it.In particular: Expand the proof by demonstrating how it works.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |