Say $G$ is a finite group with presentation $\langle S | R \rangle$ and let $C$ be the commutator subgroup of $G$. Then $\langle S | R \cup \{ sts^{-1}t^{-1} \} \rangle$ is a presentation of $G/C$.

I have been at this for a bit now and any help will be greatly appreciated. We have a map from $f:S \to G$ and the natural map $u: G \to G/C$. It is clear to me that there is a map from $f':S \to G/C$ such that $f' = u \circ f$. Now, I have been trying to find some way of putting $C$ in $F_S$. An idea that I have is to use the fact that $G/C$ is Abelian to infer that all elements of $S$ commute. Then use this information to show $F_S$ is Abelian.

But I am not sure if my approach is correct.

  • $\begingroup$ Okay, I am giving up for now. Will check again in the morning, I cannot understand my own attempt anymore. $\endgroup$ – img_teacher Aug 23 '15 at 0:04

Here is a sketch proof. There is no need to assume that $G$ is finite. Let $$Q = \langle S \mid R \cup \{ sts^{-1}t^{-1} : s,t, \in S \} \rangle.$$ Since the image in $G/C$ of each $sts^{-1}t^{-1}$ with $s,t \in S$ is trivial, there is an epimorphism $\tau:Q \to G/C$.

But since the generators of $Q$ commute with each other, $Q$ is abelian, and so, since $G/C$ is the maximal abelian quotient of $G$, $\tau$ must be an isomorphism.

  • $\begingroup$ Thank you for your answer. I am actually stumbling on showing that the said epimorphism exists. To be precise, I made use of the universal property satisfied by presentations to get a morphism from G to Q. Then I used the universal property satisfied by abelianization to get a map from Q to G/C. Would you happen to have a more lucid proof of this ? Thanks again. $\endgroup$ – img_teacher Aug 23 '15 at 15:09
  • $\begingroup$ It is also the universal property of presentations that gets you the epimorphism from $Q$ to $G/C$ because all of the relations of $Q$ lie in the kernel of the natural map from the free group on $S$ to $G/C$. $\endgroup$ – Derek Holt Aug 23 '15 at 17:22

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