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Motivation:

In trying to answer this question, I learnt that the question of determining the subgroups of finite groups is decidable.

I wrote a short argument as a (now deleted) answer:

Suppose that the general problem of finding all subgroups of any group $G$ is decidable. Then the same algorithm can be used to determine whether $G\cong\{e\}$, since if $H\le G$ implies $H=\{e\}$, then $G\cong \{e\}$. The problem is thus undecidable.

It received praise in a comment with the caveat that it only applies to infinite groups. I don't see how.

Hence the following question(s).

The Question:

Where does infinity come into play in my argument above (if at all)? Is the problem of finding all subgroups of an infinite group really undecidable?

Context:

I have only seen a handful of undecidability proofs. The above is my first attempt at one, despite working in combinatorial group theory, where there's a plethora of undecidable problems.

Searching for an answer via search engines like Google or even Approach0 is difficult due to the sheer amount of results of this kind.

I feel ill-equipped to construct nor analyse a proof either way, and was surprised my attempt above was praised.

Please help :)

Shaun
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    I'm not sure what you are asking. In the question you linked to, we have an enormous amount of information regarding the group. General Decidability questions simply don't enter into it. Keep in mind, just because a problem may be undecidable in full generality does not mean that some instance of it has to be undecidable. – lulu Oct 14 '19 at 23:30
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    Also: that question is extremely recent. If a user posted something you disagree with or that you don't understand, ask them to explain further. – lulu Oct 14 '19 at 23:31
  • I focused on the general method for calculating the subgroup structure part, @lulu; it just brought decidability to my mind. You're right: it doesn't apply. That aside, my main question is, Is the problem of finding all subgroups of an infinite group really undecidable? – Shaun Oct 14 '19 at 23:36
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    What do you mean, exactly? The usual word problem concerns a set of letters and relations. Are you saying "suppose we have the usual word problem but we are given the extra information that the group is infinite. Can we then enumerate the subgroups?" If so, then it seems the answer is "obviously not". The extra information is not helpful...were it decisive we could just add a new letter to the list and give it no relations. The new group would then be infinite and we could then solve the full problem by solving it in this case and then deleting the supernumerary letter. – lulu Oct 14 '19 at 23:40
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    here is what I'd consider a much simpler problem. Users there are asserting that even this one is undecidable and references are provided. – lulu Oct 14 '19 at 23:45
  • I mean, @lulu, that given any infinite group, is there an algorithm that will determine (in finite time?) all its subgroups? – Shaun Oct 14 '19 at 23:45
  • Ok, that's equivalent to what I wrote, so my construction would apply. – lulu Oct 14 '19 at 23:46
  • In that case, @lulu, please would you write an answer instead of leaving it in the comments? I'm inclined to accept it – Shaun Oct 14 '19 at 23:48
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    I don't think I've said much. As I mentioned, I am not convinced the problem is well posed. In practice, it is likely to come down to "how do you know the group is infinite?" I gave an example in which that information was useless..."if the generating list contains letters that do not appear in the relations." But in reality that information is likely to be more helpful. – lulu Oct 14 '19 at 23:51
  • Fair enough, @lulu. I'm not convinced it's well posed either, now that you've given some feedback. The onus is on me to make the question clear enough for an answer, not you. – Shaun Oct 14 '19 at 23:55
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    @Shaun Saying that "the problem of finding all the subgroups of a group is undecidable" doesn't even make sense. It's not even a decision problem, for a start. – verret Oct 15 '19 at 01:53
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    Second, my guess is that whoever said "it only applies to infinite groups" meant "just because some question is undecidable for groups in general, doesn't mean that it's undecidable for finite groups". (It's hard to know now that the answer is deleted.) In other words, even if your proof was correct, it would be somewhat irrelevant since the question at hand is about finite groups. – verret Oct 15 '19 at 01:56

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The problem is that in the case of finite groups deciding if something is the trivial group is decidable. A ridiculous statement, but analogous, is that in the collection of trivial groups you can decide if a group is trivial... You can do this because you already know something special about the groups.

The general problem of being able to decide if a group (say given by a presentation) is trivial (finite, hyperbolic, etc) is undecidable. Now if you have a group that is actually trivial you can verify that eventually. The problem is that you don't know when to stop if you haven't verified one of the properties: "trivial" or "not trivial".