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
\(\ds x \circ \paren {x \circ x}\) | \(=\) | \(\ds \paren {x \circ x} \circ x\) | 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.
$\blacksquare$