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Suppose you have a finite presentation of a group and you want to determine if it yields the trivial group. We know this is unsolvable in general. But say you start from the trivial group and “scramble” it by iteratively using Tietze transformations. Does this mean that once there are sufficiently many interactions introduced (via the Tietze transformations) then unsolvability arises somehow through combinatorial effects?

Using randomness on the above application of Tietze transformations is presumably necessary (select a random generator or relation, and / or continue to apply Tietze transformations for a random number of iterations). The randomness is needed as otherwise you would be able to reverse the algorithm?

The above is predicated on all presentations of a trivial group can be obtained from starting with a presentation with a singleton set of generators (identity) and singleton relator (identity) and then applying Tietze transformations. Is this valid, or does there exist a trivial group presentation that cannot be obtained this way?

Shaun
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Single Malt
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1 Answers1

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It sounds to me as though you might have misunderstood the undecidability of this problem. The problem is semi-decidable in the following sense.

If a presentation defines the trivial group, then it is always possible to prove it. Since we know that the presentation can be transformed to the presentation of the trivial group (with no generators and no relators) using Tietze transformations, you can just apply them systematically, and you will eventually achieve this. The only problem is that you cannot predict in advance how long it might take you to achieve it. More formally, the number of Tietze transformations required to effect this transformation is not in general a recursive function of the length of the input presentation.

On the other hand, if your presentation does not define the trivial group then it might not be possible to prove this, which is why the problem is unsolvable.

Derek Holt
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  • Yes I think you are right, the semi-decidability viewpoint helps understand this. This seems similar in effect to the Halting problem. Is a trivial group in its perhaps most logical presentation $\langle e|e\rangle$ (you mention no generators and no relators)? – Single Malt Mar 24 '22 at 18:15
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    I am not sure what you mean by the most logical presentation, but I would regard $\langle \mid \rangle$ as the canonical presentation of the trivial group. – Derek Holt Mar 24 '22 at 19:27