Order of Element Equals Order of Inverse

Theorem

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

Then:

$\forall x \in G: \left|{x}\right| = \left|{x^{-1}}\right|$

where $\left|{x}\right|$ denotes the order of $x$.

Proof

By Index Laws for Monoids: Negative Index, $\left({x^k}\right)^{-1} = x^{-k} = \left({x^{-1}}\right)^k$.

• Suppose $x^k = e$. Then $\left({x^{-1}}\right)^k = e$.

So $\left|{x^{-1}}\right| \le \left|{x}\right|$.

Similarly, suppose $\left({x^{-1}}\right)^k = e$.

Then $x^{-k} = e$, and so $\left({x^{-k}}\right)^{-1} = x^k = e$.

So $\left|{x}\right| \le \left|{x^{-1}}\right|$.

Thus $\left|{x}\right| = \left|{x^{-1}}\right|$.

• A similar argument shows that if $x$ is of infinite order, then so must $x^{-1}$ be.

Hence the result.

$\blacksquare$

Sources

• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.9$: Exercise $5.6$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $6.10$