I think this is not a duplicate
Let $K$ be a normal subgroup of $G$, and $H$ a normal subgroup of $K$. If $G/H$ is abelian, prove that $G/K$ and $K/H$ are both abelian.
My attempt is to set $f:G/H\to G/K$ given by $f(Hx)=Kx$ for every $x\in G$, so $f(Hx\cdot Hy)=f(H(xy))=K(xy)=Kx\cdot Ky=f(Hx)\cdot f(Hy)$, so $f$ is a homomorphism and $G/K$ is abelian (since homomorphisms preserve conmutativity)
Now $G/H$ is abelian so $H(xy)=H(yx)$ for every $x,y\in G$, but every $x,y\in K$ are also in $G$, so $K/H$ is abelian
Is the proof right? I don´t see any mistakes but I´ve never seen/thought there could be a homomorphism between two different quotient groups
Best Answer
I assume that you forgot something: in order to give a meaning to $G/H$, $H$ must be normal in $G$. I did not see that in your post.
A very nice and simple criterion for a factor group $G/N$ to be abelian is if and only if the commutator subgroup $G' \subseteq N$. I leave it to you to prove that.
In your situation: $H \unlhd G$ and $H \subseteq K \unlhd G$, and apparently $G' \subseteq H$. Since $H \subseteq K$, it follows that $G' \subseteq K$ and hence $G/K$ is abelian. Further, $K' \subseteq G' \subseteq H$, whence $K/H$ is also abelian by the criterion but now applied to $H \unlhd K$.