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Let G be a group and let G' be its commutator. Prove there is a one-to-one correspondence between the set of normal subgroups of G whose quotient is abelian and the set of all subgroup of G/G'.

I tried to use the fact that the commutator is a subgroup of any normal subgroup whose quotient is abelian but I can't seem to realize what correspondence I should be looking at. I would appreciate your help.

The allegedly duplicated question does not talk about the correspondence aforementioned nor does it help me in any way getting what I need to do understood.

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Hint: If $N \unlhd G$ then $G/N$ is abelian iff $G' \subseteq N$.

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  • $\begingroup$ Yes I know that and took it into consideration as I said. $\endgroup$ – Meitar Jan 10 '15 at 13:45
  • $\begingroup$ OK Meitar, but each subgroup of $G/G'$ can be represented as $H/G'$, where $G' \subseteq H \leq G$. $\endgroup$ – Nicky Hekster Jan 10 '15 at 14:34
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Hint. For every normal subgroup $H$ of G whose quotient is abelian define $f:H\to H/G'$.

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  • $\begingroup$ gz for answering a duplicate $\endgroup$ – Mister Benjamin Dover Jan 10 '15 at 12:17
  • $\begingroup$ I still can't seem to realize how it shows that the I can pair any normal subgroups whose quotient is abelian with a subgroup of G/G' so as to show correspondence, not isomorphism... $\endgroup$ – Meitar Jan 10 '15 at 12:24
  • $\begingroup$ @Meitar, my fault, I forgot to say that you can look at $H/G'$ like a subgroup of $G/G'$. $\endgroup$ – Jihad Jan 10 '15 at 12:26
  • $\begingroup$ How can the above be an isomorphism? If G'<H then |H|=a, but |H/G'|=[H:G']. Let H be G', then |H|=a but |H/G'| is 1, isn't that so? $\endgroup$ – Meitar Jan 10 '15 at 12:33
  • $\begingroup$ Yes, it is true. In fact $f(G')$ is a one-element subgroup. $\endgroup$ – Jihad Jan 10 '15 at 12:37

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