Powers of Group Elements/Product of Indices/Additive Notation

Theorem
Let $\struct {G, +}$ be a group whose identity is $e$.

Let $g \in G$.

Then:
 * $\forall m, n \in \Z: n \paren {m g} = \paren {m n} g = m \paren {n g}$

Proof
All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Product of Indices:


 * $\forall m, n \in \Z: g^{m n} = \paren {g^m}^n = \paren {g^n}^m$

where in this context:
 * the group product operator is $+$
 * the $n$th power of $g$ is denoted $n g$