Power of Product in Abelian Group

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Theorem

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

Then:

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


Additive Notation

This can also be written in additive notation as:

$k \left({x + y}\right) = k x + k y$


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.

$\blacksquare$


Sources