Element Commutes with Square in Group/Proof 2

Theorem

Let $\left({G, \circ}\right)$ be a group.

Let $x \in G$.

Then $x$ commutes with $x \circ x$.

Proof

By definition, a group is also a semigroup.

Thus the result Element Commutes with Square in Semigroup can be applied.

$\blacksquare$