w(G)^3=G for every non-trivial group word w and every large finite simple group G

You need to know: Group, identity element, finite group, notation $|G|$ for the number of elements in a finite group G, simple group, free group.

Background: Let $w = w(x_1,\dots,x_d)$ be a non-trivial group word, that is, a non-identity element of the free group $F_d$ on $x_1,\dots,x_d$ . For a group G, denote $w(G)$ the set of all elements $g \in G$ which can be obtained by substitution of some $g_1, g_2, \dots, g_d \in G$ into w instead of $x_1, x_2, \dots, x_d$ , respectively, and performing the group operation. For subsets $A,B$ of group G, let $A\cdot B=\{g \in G: g=a\cdot b, \, a\in A, \, b\in B\}$ . Denote $A^3=A \cdot A \cdot A$ .

The Theorem: On 24th January 2006, Aner Shalev submitted to the Annals of Mathematics a paper in which he proved that for any non-trivial group word w there exists a positive integer $N=N(w)$ such that for every finite simple group G with $|G|\geq N(w)$ we have $w(G)^3=G$ .

Short context: Given a non-trivial group word w, is there a constant $c=c(w)$ such that $w(G)^c=G$ for every large finite simple group G? This was proved for the commutator word $w=x_1^{-1}x_2^{-1}x_1x_2$ by Wilson in 1994, and for the power word $w=x_1^k$ by Martinez and Zelmanov in 1996. In 2001, Liebeck and Shalev deduced from this theorem that this is true for all non-trivial group words. However, the exact value of constant $c=c(w)$ was unknown. The Theorem proves that one can take $c=3$ , independently of w.