I have two discrete groups $G\_1$ and $G\_2$ sitting in the following exact sequences:


$1\to H\_1\to G\_1\to K\_1\to 1$ and $1\to H\_2\to G\_2\to K\_2\to 1$.


$H\_1$, $K\_1$, $H\_2$ and $K\_2$ are all non-abelian free groups of ranks $k+n$, $k$, $k+l+n$ and $k+l$ respectively. Also $k,l>1$ and $n\geq 1$. Somehow I feel that $G\_1$ and $G\_2$ are not isomorphic! May be there is some easy way to see if it is true or not.


Edit: The Hochschild-Serre spectral sequence gives the following.


$0\to H\_1(H\_1, {\Bbb Z})\_{K\_1}\to H\_1(G\_1, {\Bbb Z})\to H\_1(K\_1, {\Bbb Z})\to 0$


$0\to H\_1(H\_2, {\Bbb Z})\_{K\_2}\to H\_1(G\_2, {\Bbb Z})\to H\_1(K\_2, {\Bbb Z})\to 0$


with action of $G\_i$, $i=1,2$, is trivial on $\Bbb Z$. The action of $K\_i$ on $H\_1(H\_i, {\Bbb Z})$, $i=1,2$, giving the co-invariant $ H\_1(H\_i, {\Bbb Z})\_{K\_i}$ is mysterious!


Edit: Let $S\_1$ and $S\_2$ be two $2$-manifolds with non-abelian free fundamental groups of ranks $k$ and $k+l$ respectively. Consider the configuration space $C(S\_i)$ of $2$-tuple of ordered different points in $S\_i$, $i=1,2$. Then taking the projection to one coordinate gives a fibration $C(S\_i)\to S\_i$ with fiber $S\_i$ minus a point (Fadell-Neuwirth fibration theorem). Hence we get two exact sequences as above with $n=1$: $G\_i=\pi\_1(C(S\_i))$, $K\_i=\pi\_1(S\_i)$ and $H\_i=\pi\_1(S\_i- \{\mbox{point}\})$ for $i=1,2$.


Furthermore, consider the configuration spaces of $m$-tuple of ordered different points of $S\_i$. Then the claim is that the fundamental groups are not isomorphic. These groups are poly-free and hence the title of this thread.



Distinguishing poly-free groups
Euler characteristic is multiplicative in the setting of your exact sequences


$1\to H\_i\to G\_i\to K\_i\to 1$,


i.e. $\chi(G\_i)=\chi(H\_i)\chi(K\_i)$.


(You can see this directly by building a model for each $G\_i$ as a graph of graphs, or by more sophisticated arguments.)


In your case, this gives


$\chi(G\_1)=\chi(H\_1)\chi(K\_1)=(k-1)(k+n-1)$


while


$\chi(G\_2)=\chi(H\_2)\chi(K\_2)=(k+l-1)(k+n+l-1)$.


In particular, if $l>0$ we can see that $\chi(G\_2)>\chi(G\_1)$, which distinguishes the two groups.


