# Element of Finite Group is of Finite Order

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## 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 $1$