Power of Product of Commutative Elements in Group

Theorem

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

Let $a, b \in G$ such that $a$ and $b$ commute.

Then:

$\forall n \in \Z: \left({a \circ b}\right)^n = a^n \circ b^n$

This can be expressed in additive notation in the group $\left ({G, +}\right)$ as:

$\forall n \in \Z: n \cdot \left({a + b}\right) = \left({n \cdot a}\right) + \left({n \cdot b}\right)$

Proof

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.

$\blacksquare$

Sources

• John F. Humphreys: A Course in Group Theory (1996): $\S 12$: Proposition $12.2$