Showing that finding all subgroups of an (infinite) group is undecidable.

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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 :)

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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".