Inverse Element is Power of Order Less 1

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

\(\ds \order g\)

\(=\)

\(\ds n\)

\(\ds \leadsto \ \ \)

\(\ds g^n\)

\(=\)

\(\ds e\)

\(\ds \leadsto \ \ \)

\(\ds g^n g^{-1}\)

\(=\)

\(\ds e g^{-1}\)

\(\ds \leadsto \ \ \)

\(\ds g^{n - 1}\)

\(=\)

\(\ds g^{-1}\)

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.4$. Cyclic groups: Example $101$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Definition $3.9$: Remark $2$