# The Kervaire Conjecture

Think about your favourite non-trivial group. Now add a generator to it and then add any relation. Is the resulting group still non-trivial?

The above question is known as the Kervaire conjecture. Phrased more concretely, if $$G$$ is any non-trivial group and $$r \in G \ast \mathbb{Z}$$, is $$(G \ast \mathbb{Z})/\langle\!\langle r\rangle\!\rangle$$ again non-trivial?

This is an example of a problem which is extremely easy to state but which is still unsolved in complete generality. The best partial results so far (and that I could find by a quick search) are the following:

• It is true for finite groups, proven by Gerstenhaber and Rothaus.
• The above implies the conjecture for all residually finite groups and even all hyperlinear ones (the latter includes all sofic groups), noticed by Pestov.
• For torsion-free groups, proven by Klyachko.
• Instead of putting restrictions on the group, Klyachko and Lurye proved the conjecture under a restriction on the relation, namely that it is a proper power (i.e. $$r = w^k$$ for some $$k \ge 2$$).

Since we don’t know any non-sofic group (which is one of the major open problems in group theory), a counter-example to the Kervaire conjecture currently seems to be completely out of reach.

The origin of Kervaire’s conjecture is his paper on the classification of high-dimensional knot groups. He proved the following: Given $$n \ge 3$$. The group $$\pi$$ is isomorphic to $$\pi_1(S^{n+2} – f(S^n))$$ for some differential imbedding $$f\colon S^n \to S^{n+2}$$ if and only if $$\pi/\pi^\prime \cong \mathbb{Z}$$, the weight of $$\pi$$ is $$1$$, and $$H_2(\pi) = 0$$.

In the above statement $$\pi^\prime$$ denotes the commutator subgroup of $$\pi$$, the weight of $$\pi$$ is the smallest integer $$k$$ with the property that there exists a set of $$k$$ elements in $$\pi$$ whose normal closure equals $$\pi$$, and $$H_2(\pi)$$ is the second homology group of $$\pi$$ with integral coefficients and trivial action of $$\pi$$ on $$\mathbb{Z}$$.

At the end of his paper, Kervaire shows that the conditions $$\pi/\pi^\prime \cong \mathbb{Z}$$ and $$H_2(\pi) = 0$$ do not imply that the weight of $$\pi$$ is $$1$$, i.e. that the latter is not a redundant condition in the theorem. His example is of the form $$\pi = G \ast \mathbb{Z}$$. Hence the question whether a non-trivial free product $$G \ast \mathbb{Z}$$ can ever be of weight $$1$$ arose.