Recall that the fundamental group of a closed Riemann surface of genus $h$ has the presentation $$\Pi_h= \langle a_1, \,b_1, \ldots, a_h,\, b_h \; | \; [a_1, \, b_1]\ldots [a_h, \, b_h]=1 \rangle.$$ Assume now that we have two split short exact sequences of groups \begin{equation*} \begin{split} & 1 \longrightarrow \Pi_g \longrightarrow G_1 \longrightarrow \Pi_b \longrightarrow 1 \\ & 1 \longrightarrow \Pi_g \longrightarrow G_2 \longrightarrow \Pi_b \longrightarrow 1 \\ \end{split} \end{equation*}

Question. Is there a theoretical or computational way to check whether $G_1$ and $G_2$ are isomorphic or not?

Note that I'm talking about abstract isomorphism of the middle groups, not of isomorphism of sequences. In my specific situation (where $g=41$ and $b=2$), I know the conjugacy action of $\Pi_b$ on $\Pi_g$ in both cases, so I can obtain explicit semidirect-type presentations for $G_1$ and $G_2$ and I can feed them to GAP4.

In this way, I can check that $G_1$ and $G_2$ have the same abelianization. But the problem about their isomorphism type eludes me for the moment.

Every answer or reference to the relevant literature will be greatly appreciated.

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    $\begingroup$ Computationally, since the groups have solvable word problem, it is possible to check whether a sequence of generator images induces an isomorphism, so one can search for an isomorphism. One can also compute varionus invariants (such as the abelian invariants) and try and prove that they are not isomorphic. Both of those processes are much more difficult when there are large numbers of generators, which appears to be the case in this example. $\endgroup$
    – Derek Holt
    Commented Aug 8, 2020 at 9:17
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    $\begingroup$ @DerekHolt: Thank you for the suggestions. Actually, as I said, in my example the abelianizations (and so the abelian invariants) are the same in both cases. As you remarked, there is a large number of generators, so trying to construct an isomorphism by hand seems to be difficult (at least, for me). $\endgroup$ Commented Aug 8, 2020 at 10:51
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    $\begingroup$ Nick Salter has found examples of surface-by-surface groups that fiber in different ways. However, I don't know if these are split.msp.org/gt/2015/19-5/p10.xhtml $\endgroup$
    – Ian Agol
    Commented Aug 8, 2020 at 21:57
  • $\begingroup$ It is likely to be undecidable for the same reason that the braid factorization is undecidable, check arxiv.org/pdf/math/0511153.pdf $\endgroup$ Commented Aug 10, 2020 at 21:22

1 Answer 1


If the action of $\Pi_b$ on $\Pi_g$ via conjugation in $G$ has non-trivial kernel and infinite image in the outer automorphism group $Out(\Pi_g)$ then the extension structure for $G$ is unique. If the image of $\Pi_b$ in $Out(\Pi_g)$ is finite then $G$ is virtually a product, and there are at most two such extension structures. (This is clear when $G\cong\Pi_g\times\Pi_b$ is a product!) Otherwise, if the action is injective there are finitely many such extension structures.

These results are in (or flow from) ``A group-theoretic analogue of the Parshin-Arakelov Theorem", by F. E. A. Johnson, Archiv Math. (Basel) 63 (1994), 354-361.

Let $d=(2-2g)(2-2b)$ be the Euler characteristic of $G$. Johnson's work implies an upper bound of the order of $d^d$ on the number of such extension structures when the action is injective. On the other hand, Nick Salter showed that the number of such extension structures can be exponential in $d$.

  • $\begingroup$ This is very interesting, thank you. $\endgroup$ Commented Aug 10, 2020 at 9:04

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