Element of Finite Group is of Finite Order/Proof 1
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Theorem
In any finite group, each element has finite order.
Proof
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.
$\blacksquare$
Sources
- 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element