Finite Group Elements of Finite Order

Theorem

In any finite group, each element has finite order.

Proof

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

From Finite Semigroup Exists Idempotent Power, for every element in a finite semigroup, there is a power of that element which is idempotent.

As $G$, being a group, is also a semigroup, the same applies to $G$.

That is:

$\forall x \in G: \exists n \in \N^*: x^n \circ x^n = x^n$.

From Identity Only Idempotent Element in Group, it follows that:

$x^n \circ x^n = x^n \implies x^n = e$.

So $x$ has finite order.

$\blacksquare$

Alternative Proof

Follows as a direct corollary to the result Powers of Infinite Order Element.

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 38.3$
• John F. Humphreys: A Course in Group Theory (1996): $\S 3$: Definition $3.9$: Remark $1$