Power of Product in Abelian Group

Theorem

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

Then:

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

Additive Notation

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.

$\blacksquare$

Sources

• 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.7)$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $14$: The classification of finite abelian groups: Proposition $14.2$