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A group theory book (not in English) I'm reading states the following:

$G/G'$ is the largest abelian quotient group. In fact, every other abelian quotient group is also a quotient group of $G/G'$, due to the 3rd isomorphism theorem (for groups).

Where $G':=\langle\{a^{-1}b^{-1}ab \mid a,b\in G\}\rangle$ (the derived group).

But I don't see why this is true: Say $K$ is a normal subgroup of $G$ s.t. $G/K$ is abelian, so $G'$ is a subgroup of $K$. I understand that $K/G'$ is a quotient of $G/G'$, but it seems to state that $G/K$ is a quotient of $G/G'$?

M_N1
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    Can you expand on what you mean by "$K/G'$ is a quotient group of $G/G'$"? – Izaak van Dongen Apr 24 '24 at 14:15
  • @IzaakvanDongen Freely translated, it just says it is "a quotient", which I meant as a normal subgroup. But this might be the root of my confusion. – M_N1 Apr 24 '24 at 14:19
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    Yes, I think this is the root of your confusion. You should interpret "$G$ is a quotient of $H$" as "there is some normal subgroup $N \trianglelefteq H$ such that $G \cong H/N$", the point being that $H/N$ is "a quotient of $H$". In general "quotient" refers to "the result of dividing something by something else", not "the thing you divide by". It's like how $G \times H$ is the product of $G$ and $H$. – Izaak van Dongen Apr 24 '24 at 14:23
  • @IzaakvanDongen I see now! Great, thanks. If you'd like to write this as an answer I will accept it. – M_N1 Apr 24 '24 at 14:25
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    I'm glad to have helped. Do you now see how the third isomorphism theorem finishes the job? If so I don't mind if you write a short answer to your own question in your own words (I will give you an upvote!) But I would maybe give it 30 minutes or so in case the community decides this is a duplicate of something. – Izaak van Dongen Apr 24 '24 at 14:46
  • @IzaakvanDongen Yes I figured it out. Wrote it as an answer below, as per your suggestion. Thanks again – M_N1 Apr 24 '24 at 15:09

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Say $K$ is an abelian quotient of $G$, i.e. there is some normal subgroup $N⊴G$ such that $K≅G/N$ and $K$ is an abelian group. From the 3rd isomorphism theorem, we get that $(G/G')/(N/G')≅G/N$. So from the transitivity of isomorphisms, $(G/G')/(N/G')≅K$, which means K is a quotient $G/G'$, as requested.

Thanks to @Izaak van Dongen for guiding me in the comments!

M_N1
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