Power of Product of Commutative Elements in Group

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


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

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


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

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.

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