Element Commutes with Square in Group

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Theorem

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

Let $x \in G$.


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


Proof 1

\(\displaystyle x \circ \left({x \circ x}\right)\) \(=\) \(\displaystyle \left({x \circ x}\right) \circ x\) Group Axioms: $G1$: Associativity

$\blacksquare$


Proof 2

By definition, a group is also a semigroup.

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

$\blacksquare$