**Let G be a group and N be a normal group of G. Then (gN)^a = g^aN for all a \in \mathbb{Z}.**

**Proof.**We know that uN\cdot vN = (uv)N for all u,v \in G since N is a normal subgroup of G. This implies that gN \cdot gN = ggN = g^2N. If we apply this a, then we get

\begin{align*} (gN)^a &= gN \cdot gN \cdot gN \cdots gN \\ &= g^2N \cdot gN \cdots gN \\ & \ \ \vdots \\ &= g^aN, \end{align*}