# 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

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 ?

• $1$ is short for the trivial group $\{1\}$ or $\{e\}$. Jan 11, 2019 at 17:33
• @Randall still though how is $H/H=1$, I would of thought taking right cosets of H with elements of H would just return the group H itself ? Jan 11, 2019 at 17:34
• @can'tcauchy No How many cosets of $H$ in $H$ are there? Jan 11, 2019 at 17:35
• @can'tcauchy well, if you want, $H/H =\{H\}$, the trivial group. Jan 11, 2019 at 17:40

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 $$[a,b] \in (G/H)^{n+1}$$.
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 $$1$$.