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This question may be too general; if so, please let me know and I can try to make it more specific! Given a subgroup of $G$, $H$, with presentation $\langle S\mid R\rangle$, how do we find the presentation of the normal closure of $H$?

user1729
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Anonymous
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  • A presentation of what do you want to find? The normal closure is rarely finitely generated. – markvs Jul 09 '20 at 02:40
  • There is no "the" presentation, but rather you want "a" presentation. – user1729 Jul 09 '20 at 07:29
  • Although I answered this question, I'm now not sure exactly what you are asking. I assumed that the input was a finite generating set for $H$ (so just a finite set) and that the presentation you give is a presentation for $G$. But looking at it now, the question could be interpreted to say that the input is the presentation $\langle S\mid R\rangle$, which is a presentation for $H$, and no information is given/supposed about $G$. – user1729 Jul 09 '20 at 09:52
  • (My answer still works for both interpretations, as given a finite subset of a free group we can find a presentation for the subgroup via standard algorithms, e.g. Stallings' foldings. I just wanted to say that the question is currently unclear.) – user1729 Jul 09 '20 at 09:53

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There is no algorithm which will give you a "meaningful" output in general, even if we assume that $G$ is free. This is because non-trivial normal subgroups of free groups are finitely generated if and only if they have finite index (see here), and because it is undecidable if a finite presentation $\langle \mathbf{x} \mid\mathbf{r}\rangle$ defines a finite group (as "being finite" is a Markov property). Therefore, in the free group $F(\mathbf{x})$ we cannot determine whether it not the normal closure of the set $\mathbf{r}$ has finite index or not, and hence if it is finitely generated or not.

I used the word "meaningful" above as there may exist an algorithm which will output a description of some generating set of the normal closure of $\mathbf{r}$, but then by the above it is impossible to use this description to find a basis for the subgroup. So I think the above captures the essence of the question :-)

user1729
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