Inverse Element is Power of Order Less 1

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Theorem

Let $G$ be a group whose identity is $e$.

Let $g \in G$ be of finite order.

Then:

$\order g = n \implies g^{n - 1} = g^{-1}$


Proof

\(\displaystyle \order g\) \(=\) \(\displaystyle n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle g^n\) \(=\) \(\displaystyle e\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle g^n g^{-1}\) \(=\) \(\displaystyle e g^{-1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle g^{n - 1}\) \(=\) \(\displaystyle g^{-1}\)

$\blacksquare$


Sources