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

  • $\begingroup$ Can you tell from where did you get this question? $\endgroup$
    – Error 404
    Feb 24, 2017 at 16:57
  • 2
    $\begingroup$ Your proof looks good, though to be completely rigorous, you need to show that your map $f$ is well-defined (i.e. independent of the coset representative chosen) - why is this? Alternatively, it's not too hard to show that any subgroup of an abelian group is abelian, as is any quotient. Since $K/H$ is a subgroup of the abelian group $G/H$, it is also abelian; and $G/K \cong (G/H)/(K/H)$ by the third isomorphism theorem, so it must also be abelian. $\endgroup$ Feb 24, 2017 at 16:58
  • $\begingroup$ Vikrant Desai, from "A book of abstract algebra" (by Charles C. Pinter) $\endgroup$ Feb 24, 2017 at 16:59
  • $\begingroup$ Alex Wertheim I wasn't sure I had to check that, but your proof is much easier, thank you $\endgroup$ Feb 24, 2017 at 17:29
  • $\begingroup$ I have a problem here. The group $G = (C_4 \times C_4) \rtimes C_9$ has a normal subgroup $K = C_{12} \times C_4$ which in turn has a normal subgroup $H = C_{12}\times C_2$, but $H$ is not normal in $G$ how can you speak about $G/H$? $\endgroup$ Feb 25, 2017 at 0:53

1 Answer 1


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 $H \unlhd K$.

  • 1
    $\begingroup$ For the record, the proof that G/H is abelian iff all commutators are in H is included in Pinter. Page 152! $\endgroup$
    – K.Reeves
    Apr 11, 2017 at 3:50

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