I'm having trouble coming to grips with the fact that when dealing with the composition of cosets in a factor group, it will be true that $g^iH = (gH)^i$. If I think about the rule that's actually given to me:
$$ (aH)(bH)=abH $$
I'm quite happy with that. If we're dealing with cyclic groups, I understand that given a cyclic group $G$ and a subgroup $H$ then the factor group should (more or less) be defind as the following:
$$ G/H=\{g^iH \mid i \in \mathbb{Z} \} $$
I'm good with that also. I just can't make whatever leap is necessary to understand the implication that $g^iH = (gH)^i$. All of the insights I have seen are some derivative of, "it's given by the definition," which is not good enough.
For context, I am working through trying to prove that a factor group of a cyclic group is cyclic, and this seems to be what I need to make everything work.
I suspect that I am not thinking abstractly enough. My brain keeps telling me to turn $(gH)^i$ into $g^iH^i$ which I don't think is a thing. Apparently it's an easy enough induction proof, but I haven't seen one nor have I been able to construct one for the same reason.
Any insights will be appreciated. Apologies if I've missed something blatantly obvious.
I will only prove the case for $i=2$. Prove the other cases by induction.
Let $a \in (gH)^2 = (gH)(gH)$, i.e. there is $b,c \in H$ such that $a = gbgc$. Then, $a = g^2 bc$ (cyclic group is abelian) and $bc \in H$, so $a \in g^2 H$.
Let $a \in g^2 H$, i.e. there is $b \in H$ such that $a=g^2b$, so $a = gegb$, so $a \in (gH)(gH)$.