Power of Product of Commutative Elements in Group

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Theorem

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

Let $a, b \in G$.


Then:

$a \circ b = b \circ a \iff \forall n \in \Z: \paren {a \circ b}^n = a^n \circ b^n$


That is:

$a$ and $b$ commute

if and only if:

the product of their powers equals the power of their product:


This can be expressed in additive notation in the group $\struct {G, +}$ as:

$a + b = b + a \iff \forall n \in \Z: n \cdot \paren {a + b} = \paren {n \cdot a} + \paren {n \cdot b}$


Proof

Necessary Condition

Let $a \circ b = b \circ a$.

By definition, all elements of a group are invertible.

Therefore the results in Power of Product of Commutative Elements in Monoid can be applied directly.

$\Box$


Sufficient Condition

If $\paren {a \circ b}^n = a^n \circ b^n$ for all $n$, then it certainly holds for $n = 2$:

\(\displaystyle \paren {a \circ b}^2\) \(=\) \(\displaystyle a^2 \circ b^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {a \circ b} \circ \paren {a \circ b}\) \(=\) \(\displaystyle \paren {a \circ a} \circ \paren {b \circ b}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {a^{-1} \circ a} \circ b \circ a \circ \paren {b \circ b^{-1} }\) \(=\) \(\displaystyle \paren {a^{-1} \circ a} \circ a \circ b \circ \paren {b \circ b^{-1} }\) Group Axiom $G \, 1$: Associativity
\(\displaystyle \leadsto \ \ \) \(\displaystyle e \circ b \circ a \circ e\) \(=\) \(\displaystyle e \circ a \circ b \circ e\) Group Axiom $G \, 3$: Inverses
\(\displaystyle \leadsto \ \ \) \(\displaystyle b \circ a\) \(=\) \(\displaystyle a \circ b\) Group Axiom $G \, 2$: Identity

$\blacksquare$


Sources