Element of Finite Group is of Finite Order

Theorem

In any finite group, each element has finite order.

Proof 1

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

From Element has Idempotent Power in Finite Semigroup, 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:

$\forall x \in G: \exists n \in \N_{>0}: x^n \circ x^n = x^n$

From Identity is only Idempotent Element in Group, it follows that:

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

So $x$ has finite order.

$\blacksquare$

Proof 2

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

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $5$: Subgroups: Exercise $15$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Definition $3.9$: Remark $1$