For G/N, (gN)^a = g^aN for all integers a

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
(gN)^a &= gN \cdot gN \cdot gN \cdots gN \\
&= g^2N \cdot gN \cdots gN \\
&  \ \ \vdots \\
&= g^aN,
which proves the statement.

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