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.

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