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Let $G$ be a finitely presented group with a subgroup $H$(if it helps we can assume that $H$ is finitely presented as well.) Is there any method in order to check that whether $G$ splits over the subgroup $H$ or not?

Bobby
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  • I don't know exactly, but if you know enough about $G$ and $H$ it seems like you could attempt to create a Bass-Serre tree action by "knowing" edge stabilizers and try to figure out what the vertex stabilizers would have to be. I am guessing you need a lot more than just finitely presented get an actual algorithm, assuming something like what I am thinking could be carried out. –  Sep 18 '18 at 17:13

2 Answers2

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At least, there's no algorithm, say when $G$ is input by a presentation and when $H$ is input by giving generators.

Indeed, if there were such an algorithm, then we would deduce an algorithm to determine whether $G$ splits over $\{1\}$. Applying this algorithm to the free product $G\ast G$ (inputting a finite presentation of $G$), we deduce an algorithm solving the triviality problem (says yes iff $G\neq\{1\}$). It is classical that there is no such algorithm.

YCor
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There is an unsolved conjecture by Kropholler and Roller concerning how to check a slightly weaker condition, namely whether $G$ splits over a subgroup that is commensurable to $H$. Since that conjecture is still unresolved, your stronger question is also unresolved.

Lee Mosher
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  • Neither seem to imply the other, unless I am missing something. It seems like determining if $G$ splits over a subgroup commensurable to $H$ doesn't necessarily mean you can determine it splits over $H$. If you can determine whether or not $H$ splits $G$, in the not case it doesn't seem to imply you could determine if something commensurable would split it. –  Sep 19 '18 at 22:57
  • Well, the subgroup $H$ is commensurable to itself, so if $G$ splits over $H$ then it splits over some subgroup commensurable to $H$. – Lee Mosher Sep 19 '18 at 23:45
  • Yes but if you can determine $H$ doesn't split you may not know if something commensurable would make it split. Even being able to determine if a group splits up to taking a commensurable subgroup doesn't seem to say that $H$-almost invariant sets $X$ with $HXH=X$ give splittings. –  Sep 20 '18 at 00:20
  • All I am asserting in my answer is that the statement "$G$ splits over $H$" implies the statement "There exists a subgroup $K$ of $G$ such that $K$ is commensurable to $H$ and such that $G$ splits over $K$" (and I am asserting that the latter statement is a well known unsolved conjecture). My answer leaves out any assertions about the converse of that implication; you are, of course, correct that the converse is false. – Lee Mosher Sep 20 '18 at 13:37
  • What I am trying to say is that I don't see how the OP's question would answer this Kropholler and Roller conjecture. For example if you came up with a way to determine if a 3-manifold(with a few adjectives) fibered over a circle, that wouldn't necessarily answer the virtual fibered conjecture, all it says is that if you fiber over a circle you satisfy the virtual fibered conjecture(now theorem). –  Sep 20 '18 at 21:03