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

Question

For the quotient group $G/N$, $(gN)^a=g^a N$ for all integers $a$.

How do I prove this? Do I proceed with induction? In my class, we've never defined what it means to raise a coset to a power. How is $(gN)^a$ to be interpreted?

Answer 1

Since this is a quotient group, we have that $N$ is a normal subgroup of G.

Let $gN \in G/N$

We know that $(aN)(bN) = (ab)N$ since $N$ is normal.

Then for $(gN)^2 \implies (gN)(gN)=(gg)N=g^2 N$

Then we can generalize for $a$ that $(gN)^a \implies (gN)(gN)...(gN)$ given $a$ times

$\implies (gN)^a = g^a N$

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Follow up:

Inverses don't have an important part to play in this proof, although clearly $G/N$ is a group and so $g^{-1} N$ exists.

$(g^{-1} N)(g^{-1} N) = (g^{-1} g^{-1}) N = (g^{-1})^2 N$

So you can see

$(g^{-1} N)^a = (g^{-1})^a N$