# Exact sequence induces exact sequence

Consider exact sequence $N\xrightarrow{f} G\xrightarrow{g} Q\rightarrow 0$

Question is to prove that this gives exact sequence $N/[G,N]\xrightarrow{\bar{f}} G/[G,G]\xrightarrow{\bar{g}} Q/[Q,Q]\rightarrow 0$.

I could see that $N\rightarrow G\rightarrow Q$ induces Morphisms $$N/[G,N]\xrightarrow{\bar{f}} G/[G,G]\xrightarrow{\bar{g}} Q/[Q,Q]$$

and that $\bar{g}$ is surjective...

I also proved that $Im(\bar{f})\subset Ker(\bar{g})$

Only thing that i have to prove is the other way inclusion....

Let $[m]\in G/[G,G]$ such that $\bar{g}[m]=[0]$ i.e., $g(m)\in[Q,Q]$

I am not sure what to do here after... I have to use that $Ker(g)\subset Im(f)$ but i am not very sure how to get an elmenet in $Ker(g)$ with $g(m)\in [Q,Q]$..

This question is from A Course in Homological Algebra by Hilton Stammbach...

It is suggested that i am not supposed to use any homological algebra technique...

That is the only reason for adding homological algebra tag...

• What is ${[}\cdot, \cdot{]}$? – Arthur Nov 21 '14 at 12:35
• what is $[G, N]$? – Krish Nov 21 '14 at 12:54
• $[G,G]$ is commutator subgroup of $G$... $[G,N]=\{\langle gng^{-1}n^{-1}\rangle : g\in G n\in N\}$ @Arthur – user87543 Nov 21 '14 at 12:55
• So $gng^{-1}n^{-1}\in N$? How would that work? – Arthur Nov 21 '14 at 15:22
• @Arthur : oh... yes.. I don't know why it was written like that in that text :O – user87543 Nov 21 '14 at 16:18

In the following proof I will use that the functor $G \mapsto G/[G,G]$ from groups to abelian groups is left adjoint to the forgetful functor. If $T$ is an abelian group, then
$\hom(Q/[Q,Q],T) \cong \hom(Q,T) \cong \{h \in \hom(G,T) : N \to G \to T \text{ trivial}\}\\ \cong \{h \in \hom(G/[G,G],T) : N \to G \to G/[G,G] \to T \text{ trivial}\}\\ = \{h \in \hom(G/[G,G],T) : N/[G,N] \to G/[G,G] \to T \text{ trivial}\}.$
This shows that $N/[G,N] \to G/[G,G] \to Q/[Q,Q] \to 0$ is a cokernel diagram, i.e. it is exact.
If $[x] \in G/[G,G]$ is mapped to $0 \in Q/[G,G]$, this means that $g(x) \in [Q,Q]$, so that we may write $g(x)$ as a product of commutators $p_i q_i p_i^{-1} q_i^{-1}$. Find preimages $y_i \in G$ of $p_i$ and $z_i \in G$ of $q_i$. If $u \in G$ denotes the product of the commutators $y_i z_i y_i^{-1} z_i^{-1}$, then $g(x)=g(u)$, so that there is some $n \in N$ with $x = f(n) u$. Thus, $[x]=[f(n)]$ lies in the image of $N/[G,N] \to G/[G,G]$.
• Sir, i tried something similar but then i do not see why it is $N/[G,N]$ and not $N/[N,N]$.. – user87543 Nov 21 '14 at 14:45
• It doesn't really matter, the sequence becomes exact with both groups. The reason is that both are quotients of $N$, and only the image in $G/[G,G]$ is of interest. – Martin Brandenburg Nov 21 '14 at 14:53