Powers of Group Element Commute

Theorem

Let $\struct {G, \circ}$ be a group.

Let $g \in G$.

Let $m, n \in \N_{>0}$.

Then:

$\forall m, n \in \N_{>0}: g^n \circ g^m = g^m \circ g^n$


Proof

By definition, a group is also a semigroup.

The result follows as a special case of Powers of Semigroup Element Commute

$\blacksquare$


Sources

  • 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.12)$
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