Suppose for an arbitrary group word $w$ ower the alphabet of $n$ symbols $\mathfrak{U_w}$ is a variety of all groups $G$, that satisfy an identity $\forall a_1, … , a_n \in G$ $w(a_1, … , a_n) = e$. Is it true, that for any group word $w$ there exists a positive real number $\epsilon (w) > 0$, such that any finite group $G$ is in $\mathfrak{U_w}$ iff $$\frac{\lvert\{(a_1, … , a_n) \in G^n : w(a_1, … , a_n) = e\}\rvert}{{|G|}^n} > 1 – \epsilon(w)?$$

How did this question arise? There is a widely known theorem proved by P. Erdős and P. Turán that states:

A finite group $G$ is abelian iff $$\frac{|\{(a, b) \in G^2 : [a, b] = e\}|}{{|G|}^2} > \frac{5}{8}.$$

This theorem can be rephrased using aforementioned terminology as $\epsilon([a, b]) = \frac{3}{8}$.

There also is a generalisation of this theorem, stating that a finite group $G$ is nilpotent of class $n$ iff $$\frac{|\{(a_0, a_1, … , a_n) \in G^{n + 1} : [ … [[a_0, a_1], a_2]… a_n] = e\}|}{{|G|}^{n + 1}} > 1 – \frac{3}{2^{n + 2}},$$ thus making $\epsilon([ … [[a_0, a_1], a_2]… a_n]) = \frac{3}{2^{n + 2}}$.

However, I have never seen similar statements about other

one-wordvarieties being proved or disproved, despite such question seeming quite natural . . .Actually, I doubt that the conjecture in the main part of question is true. However, I failed to find any counterexamples myself.

**Answer**

**Attribution***Source : Link , Question Author : Chain Markov , Answer Author : Community*