**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.4**

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

Solution: 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 $k$.