Two questions regarding this proof from my notes which shows that if $H$ is normal in $G$ and $G$ is solvable then $\tfrac{G}{H}$ is solvable

Question

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 ?

Answer 1

i) We're allowed to say it if we can prove it (as is the case with everything in mathematics). A proof could go as follows : by induction: it's clear for $n=1$; now if it's true for $n$, let $x\in (G/H)^{n+1} = D((G/H)^n)$. Then $x = [a,b]$ for some $a,b\in (G/H)^n$. By the induction hypothesis, $a$ and $b$ can be written as the class mod $H$ of $t,s \in G^n$. So $x$ is the class mod $H$ of $[t,s]\in G^{n+1}$, so $x\in G^{n+1}H/H$.

Conversely, let $x \in G^{n+1}H/H$. Then $x$ is the class mod $H$ of some $y\in G^{n+1}$. But then $y=[a,b]$ for some $a,b\in G^n$. By the induction hypothesis, the classes of $a,b$ mod $H$ are in $(G/H)^n$, so the class of $[a,b]$ is in $(G/H)^{n+1}$, so $x=$ the class of $y= $ the class of $[a,b] \in (G/H)^{n+1}$.

The result follows.

ii) There is only one class of $H$ mod $H$, as is the case for any group $G$. Indeed, if $x,y\in H$, then $xH= H = yH$.Therefore $H/H$ is a group with one element, there is only one up to isomorphism and we denote it by $1$.