# homomorphism inducing monomorphism on some quotient group

Let $f:G\rightarrow H$ be a group homomorphism such that $f_* :G_{ab}\rightarrow H_{ab}$ is an isomorphism and that $f_* : H_2(G)\rightarrow H_2(H)$ is an epimorphism.

Question is to prove that this induced a monomorphism $$f:\frac{G}{\bigcap_{n=0}^{\infty}G_n}\rightarrow \frac{H}{\bigcap_{n=0}^{\infty}H_n}$$

Define the abelianization of a group $G$ to be the quotient group $G_{ab} := G/[G,G]$, where $[G,G]$ is the commutator subgroup. $G_n$ is the lower central series defined inductively as $G_1=[G,G]; G_{n+1}=[G,G_n]$

I see that $f(G_n)\subset H_n$ so we have induced map $f: G/G_n\rightarrow H/H_n$ and for similar reasons we have well defined map

$$f:\frac{G}{\bigcap_{n=1}^{\infty}G_n}\rightarrow \frac{H}{\bigcap_{n=1}^{\infty}H_n}$$

Now the question is how do you show that this is a monomorphism..

Let $[x]\in \frac{G}{\bigcap_{n=1}^{\infty}G_n}$ such that $[f(x)]=0\in \frac{H}{\bigcap_{n=1}^{\infty}H_n}$ i.e., $f(x)\in \bigcap_{n=1}^{\infty}H_n$

As $f(x)\in \bigcap_{n=1}^{\infty}H_n$ we have in particular $f(x)\in H_1$.

As $f_* : G/[G,G]\rightarrow H/[H,H]$ is an isomorphism and $f(x)\in [H,H]$ we have $x\in [G,G]$

So, there is some hope.. If $f(x)\in H_1$ we have $x\in G_1$...

As $f(x)\in H_2=[H,H_1]$.. Taking for granted all better possibilities we have $f(x)=h(aba^{-1}b^{-1})h^{-1}(aba^{-1}b^{-1})^{-1}$... I am not sure where to go from this...

I am not sure how to use other details of that isomorphism to conclude $x\in G_n$ for all $n\in \mathbb{N}$

• This can't be right! We could have $H = G/[G,G]$ for example. – Derek Holt Nov 22 '14 at 11:02
• i am not sure... For reference this is an exercise from A course in homological algebra by hilton stammbach chapter 6, exercise 9.1... @DerekHolt – user87543 Nov 22 '14 at 11:04
• But as it stands it is clearly wrong! Are you sure you haven't missed out a hypothesis? – Derek Holt Nov 22 '14 at 11:05
• I intentionally did not write one hypothesis as i was sure that this is not related.. I have added it now.. please check it.. @DerekHolt – user87543 Nov 22 '14 at 11:06
• Ah right! That makes it harder! It's late at night here, so I'll have to leave it for today. – Derek Holt Nov 22 '14 at 11:09

By Theorem 9.1 (Chap. VI) in Hilton-Stammbach's A Course in Homological Algebra, $f$ induces for each $n$ an isomrphism $$f_n: G/G_n \to H/H_n.$$
Now let $x\in G$ represent an element from the kernel of $G/\bigcap_nG_n \to H/\bigcap_n H_n$. Hence $f(x)\in \bigcap_nH_n\subseteq H_n$. In particular $f_n(\bar{x})=\bar{1}$ and, by the theorem, $\bar{x}=\bar{1}\in G/G_n$. Thus $x\in G_n$. This holds for all $n$, so $x \in \bigcap_nG_n$ and we are done.