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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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    $\begingroup$ Have you checked their abelianisations? $\endgroup$
    – HJRW
    Commented Dec 9, 2021 at 8:25
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    $\begingroup$ These are "meta-free" groups. Poly-free would allow iterated extensions. $\endgroup$
    – YCor
    Commented Dec 9, 2021 at 9:28
  • $\begingroup$ Presumably you are given the groups in some explicit way -- for instance, via homomorphisms $H_i\to\mathrm{Aut}(K_i)$. If so, you can compute a presentation for the $G_i$, and then abelianise these to compute the abelianisations of the groups. If this fails to distinguish them, you can try to count homomorphisms from $G_i$ to some of your favourite finite groups. This is often a practical way to disitinguish pairs of groups. $\endgroup$
    – HJRW
    Commented Dec 9, 2021 at 14:05
  • $\begingroup$ Maybe some further remarks are in order about why this problem is difficult in general. The best-case scenario one might hope for is that the subgroups $H_i$ are somehow canonical, and perhaps the isomorphism class of the group $G_i$ determines the (outer) action of $K_i$ on $H_i$. Unfortunately this is very far from true. To see this, consider any fibred hyperbolic link complement $M$. (I think the Borromean rings give an example.) Thurston showed that the set of fibrations occupy an open cone in $H^1(M,\mathbb{Z})$, and so $M$ fibers in many different ways. $\endgroup$
    – HJRW
    Commented Dec 9, 2021 at 17:18
  • $\begingroup$ Thanks a lot HJRW for your comments. As the kernels and the quotients are both non-abelian free, I thought the situation will be more rigid compared to the fibration over a circle case, as in Thurston's result. I am editing my question with an explicit example, it is also induced by a fibration. $\endgroup$
    – RKS
    Commented Dec 10, 2021 at 5:18

1 Answer 1

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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.

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  • $\begingroup$ Thanks a lot! I checked in Kenneth Brown's book Cohomology of Groups' Prop. 7.3 (d). Finitely generated free groups are of finite homological types, for which the above multiplicative property of Euler characteristic holds and furthermore the middle group also becomes finite homological type. Hence induction argument gives some conclusion for poly-free groups as well. In your writing you mean graph of groups'. $\endgroup$
    – RKS
    Commented Dec 10, 2021 at 14:24
  • $\begingroup$ Well, you can think of it as a graph of groups, but in this case I think of it as a graph of graphs. :) But yes, the general statement should be in Brown's book. $\endgroup$
    – HJRW
    Commented Dec 10, 2021 at 14:25

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