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:

$\forall m, n \in \Z: n \cdot \paren {m \cdot g} = \paren {m \times n} \cdot g = m \cdot \paren {n \cdot g}$

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 operation is $+$
the $n$th power of $g$ is denoted $n \cdot g$

$\blacksquare$