# Element Commutes with Square in Group

## Theorem

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

Let $x \in G$.

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

## Proof 1

 $\ds x \circ \paren {x \circ x}$ $=$ $\ds \paren {x \circ x} \circ x$ Group Axiom $\text G 1$: Associativity

$\blacksquare$

## Proof 2

By definition, a group is also a semigroup.

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

$\blacksquare$