0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

Hint. For every normal subgroup $H$ of G whose quotient is abelian define $f:H\to H/G'$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.