# Does the union of all finite groups yield a complete knot invariant for prime knots?

It is established in Whitten - Knot complements and groups together with the Gordon-Luecke theorem (that knot complements determine knot type) that the type of a prime knot is determined by the isomorphy type of its knot group.

In the book Charles Livingston - Knot Theory, the author uses surjective homomorphisms from knot groups into finite groups as knot invariants (i.e., two knot groups are nonisomorphic if one of them can be mapped onto a certain finite group and the other one can't). My question is:

If two prime knots are distinct, does that mean there is a finite group such that exactly one of them can be mapped surjectively into it?

or:

Do all finite groups combined yield (as described above) a complete knot invariant for prime knots?

## 1 Answer

Though it is not completely obvious, it turns out that if $$G_1$$ and $$G_2$$ are finitely generated groups that surject onto the same set of finite groups, then the profinite completions of $$G_1$$ and $$G_2$$ are isomorphic (you might expect that you need some kind of multiplicities here, but they are actually not needed!). So what you're asking is equivalent to asking if prime knots are determined by the profinite completions of their fundamental groups.

I don't know the answer to this question, but there is a huge literature on these kinds of profinite rigidity questions for 3-manifold groups. For a recent survey of what is known, I recommend Alan Reid's ICM address, available here. See especially Section 4. By the way, the result I allude to in the first paragraph is Theorem 2.2 in this survey.

• That's a great summary! In particular, I think this question is still wide open, though maybe one needs to restrict to hyperbolic or torus knots: the comments about non-trivial JSJ decomposition indicate that perhaps the question is false for satellite knots. – Mike Miller Dec 6 at 22:24
• This may not be entirely relevant. Serre has an example of two 4 manifolds which are not homeomorphic for which the profinite completions of the fundamental groups are isomorphic. – Kapil Dec 7 at 3:01
• @Kapil: Isn't that much easier? Since all finitely presented groups are fundamental groups of compact 4-manifolds, this is just equivalent to the fact that there exist non-isomorphic finitely presentable groups with isomorphic profinite completions. – Andy Putman Dec 7 at 3:24
• True. I was thinking of Serre's example which is of an algebraic variety over a number field $K$ which has non-homeomorphic complex points under distinct embeddings of $K$ in $\mathbb{C}$. That is much more than what is required here. – Kapil Dec 7 at 12:33