## Powers in a quotient group

Let $G$ be a group and $N$ a normal subgroup of $G$. Show that for all $g \in G$ and $k \in \mathbb{Z}$, $(gN)^k = (g^k)N$.

First we show that the conclusion holds for nonnegative $k$ by induction. Note that $(gN)^0 = N = (g^0)N$. Now suppose the conclusion holds for $k \geq 0$; then $(gN)^{k+1} = (gN)(gN)^k$ $= (gN)(g^kN)$ $= (g^{k+1})N$. So the conclusion holds for nonnegative $k$ by induction.

Now suppose $k < 0$. Then $(gN)^k = ((gN)^{-k})^{-1} = (g^{-k}N)^{-1}$ $= (g^k)N$. Thus the conclusion holds for all integers $k$.