# Power of Product of Commutative Elements in Group

## Theorem

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

Let $a, b \in G$.

Then:

$a \circ b = b \circ a \iff \forall n \in \Z: \paren {a \circ b}^n = a^n \circ b^n$

That is:

$a$ and $b$ commute
the product of their powers equals the power of their product:

This can be expressed in additive notation in the group $\struct {G, +}$ as:

$a + b = b + a \iff \forall n \in \Z: n \cdot \paren {a + b} = \paren {n \cdot a} + \paren {n \cdot b}$

## Proof

### Necessary Condition

Let $a \circ b = b \circ a$.

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.

$\Box$

### Sufficient Condition

If $\paren {a \circ b}^n = a^n \circ b^n$ for all $n$, then it certainly holds for $n = 2$:

 $\ds \paren {a \circ b}^2$ $=$ $\ds a^2 \circ b^2$ $\ds \leadsto \ \$ $\ds \paren {a \circ b} \circ \paren {a \circ b}$ $=$ $\ds \paren {a \circ a} \circ \paren {b \circ b}$ $\ds \leadsto \ \$ $\ds \paren {a^{-1} \circ a} \circ b \circ a \circ \paren {b \circ b^{-1} }$ $=$ $\ds \paren {a^{-1} \circ a} \circ a \circ b \circ \paren {b \circ b^{-1} }$ Group Axiom $\text G 1$: Associativity $\ds \leadsto \ \$ $\ds e \circ b \circ a \circ e$ $=$ $\ds e \circ a \circ b \circ e$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds b \circ a$ $=$ $\ds a \circ b$ Group Axiom $\text G 2$: Existence of Identity Element

$\blacksquare$