# Power of Product in Abelian Group

## Theorem

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

Then:

$\forall x, y \in G: \forall k \in \Z: \paren {x \circ y}^k = x^k \circ y^k$

This can also be written in additive notation as:

$\forall x, y \in G: \forall k \in \Z: k \cdot \paren {x + y} = \paren {k \cdot x} + \paren {k \cdot y}$

## Proof

By definition of abelian group, $x$ and $y$ commute.

That is:

$x \circ y = y \circ x$

The result follows from Power of Product of Commutative Elements in Group.

$\blacksquare$