Consider the following theorem :
Let $G$ be a group. If $H$ is normal in $G$, and $G$ is solvable then $\tfrac{G}{H}$ is solvable.
According to my lecture notes the proof proceeds as follows:
If $H$ is normal in G and $G^n=1$ for some $n\in \Bbb N$.
$\Rightarrow (\tfrac{G}{H})^n=\tfrac{G^nH}{H}=\tfrac{H}{H}=1$, which shows that $\tfrac{G}{H}$ is solvable.
My questions:
i) Are we allowed to say that $(\tfrac{G}{H})^n=\tfrac{G^nH}{H}$ Because H is a normal subgroup and $G^n$ ( as it is contained in H) must be a subgroup of the normaliser $\Rightarrow G=G^nH$. It reminds me of the fratini argument but then that is only true for P-groups and even still $G^n$ is only a subgroup of the normaliser not the whole thing. Is there a better/correct reason for why this is true?
ii) I do not see how $\tfrac{H}{H}=1$ sure this would be true if they were numbers but it seems non-sensical if they are groups. Could anyone please explain this ?