Properties of a non-sofic group This question is, essentially, a comment of Mark Sapir. I think it deserves to be a question.
A countable, discrete group $\Gamma$ is sofic if for every $\epsilon>0$ and finite subset $F$ of $\Gamma$ there exists an $(\epsilon,F)$-almost action of $\Gamma$. See, for example Theorem 3.5 of the nice survey of Pestov http://arxiv.org/PS_cache/arxiv/pdf/0804/0804.3968v8.pdf.
Gromov asked whether all countable discrete groups are sofic. It is now widely believed that there should be a counterexample to this.
Since most groups are sofic, it would be useful to have a collection of properties that would imply that a group is not sofic...so one can then construct a beast having such properties.

What are some abstract properties of $\Gamma$ that would imply $\Gamma$ is not sofic?

An open question of Nate Brown asks whether all one-relator groups are sofic. I'd be interested to know what properties of a one-relator group $\Gamma$ would imply that $\Gamma$ is not sofic.
 A: Sofic groups fulfill determinant conjecture. 
This implies in particular that there exists a natural constant $c$ such that given a matrix $M$ over the integral  group ring of a given sofic group $G$, we have that
$$
|tr_{vN} \exp(-cM) - \dim_{vN}\ker M| < \frac{1}{3}.
$$
This can be used to show that some problems about the group are decidable. Suppose a group $G$ is torsion-free, has decidable word problem, fulfills Atiyah conjecture, and is sofic. Then there is an algortihm which decides whether a given matrix $M$ over the integral group ring has non-trivial kernel, as an operator on $[l^2(G)]^{\dim M}$. 
Indeed, given $M$ it's easy to bound its $l^2$ norm and based on this to decide how many terms in $tr_{vN}\exp(-cM)$ have to be computed in order to be less than $\frac{1}{6}$ from the actual value of $tr_{vN} \exp(-cM)$. Call this approximation $a$ (it can be computed since the word problem is decidable). Now, because $G$ is torsion free and fulfills Atiyah conjecture, we know that $\dim_{vN}\ker M$ is an integer, and it's equal to $0$ iff $M$ has trivial kernel. So $M$ kas trivial kernel if and only if $a<\frac{1}{2}$
Similar algorithm works if a group has bounded torsion, since $\frac{1}{3}$ in the first equation can be exchanged with any postivie real number. I seem to have read that there exist Tarski monsters with decidable word problem. That means that in principle :-) one could try to show that there's no such algorithm for these Tarski monsters and arrive at the conclusion that either these monsters are non-sofic or they don't fulfill Atiyah conjecture.
A: Let $\Gamma$ be a sofic group. Gabor Elek and Endre Szabo showed here, that for any field $k$ and $a,b \in k[\Gamma]$ with $ab=1$ one has $ba=1$. Hence, coming up with a cleverly chosen group where this fails would provide a counterexample. Note that $k=\mathbb C$ is not a good start since Kaplansky showed long ago that the implication holds for fields of characteristic zero. However, for $k= GF(2)$ one might be lucky.
Let us consider $k=GF(2)$ for now. One strategy could be to start with $a = \sum_{g \in F} g$ and $b = \sum_{h \in K} h$ for some finite sets $F,K \subset G$. If $ab=1$, then a number of relations must hold: We quickly convince ourselves that $F$ and $K$ must have an odd number of elements and there exists a self-matching of the set $F \times K \setminus (f,k)$ such that matched pairs $(f',k') \sim (f'',k'')$ satisfy $f'k' = f''k''$ and $f=k^{-1}$ for the special unmatched pair. You can now turn everything around and start with an abstract group with generators $F \cup K$ and relations as above coming from an arbitrarily chosen self-matching. In the group ring of this abstract group, we will have $ab=1$, but why do we have $ba=1$? I was working on this for a while but could not come up with a counterexample. On the other hand, if $F$ and $K$ are large, I cannot believe that $ba=1$ will always hold.
