Powers distribute over an operation if and only if the group is abelian.

$(a b)^{n} = a^{n} b^{n}$ if and only if $G$ is an abelian group.

Proof

Suppose $G$ is abelian.
Then

\begin{aligned} (a b)^{n} & = \underset{n}{\underset{⏟}{(a b) (a b) \dots (a b)}} \\ = a b a b \dots a b \\ = a a a \dots b b b \\ = a^{n} b^{n} \end{aligned}

Conversely, suppose $(a b)^{n} = a^{n} b^{n}$ for all $n > 0$ .
Let $n = 2 :$

a b a b = a a b b

b a = a b

so $G$ is abelian.

QED