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.

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