w_1(A_n)w_2(A_n)=A_n for every non-trivial group words w_1,w_2 and every large alternating group A_n

You need to know: Group, identity element, finite group, simple group, free group, alternating group $A_n$ of degree n.

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 $A\cdot B=\{g \in G: g=a\cdot b, \, a\in A, \, b\in B\}$ .

The Theorem: On 11th January 2007, Michael Larsen and Aner Shalev submitted to arxiv a paper in which they proved that (i) for each pair of non-trivial words $w_1$ , $w_2$ there exists $N =N(w_1, w_2)$ such that for all integers $n \geq N$ we have $w_1(A_n)w_2(A_n) = A_n$ , and (ii) for every triple of non-trivial group words $w_1, w_2, w_3$ , there exists $N =N(w_1, w_2,w_3)$ such that $w_1(G)w_2(G)w_3(G) = G$ for every finite simple group with at least N elements.

Short context: In an earlier paper, Shalev proved that $w^3(G)=G$ for every non-trivial group word w, and every sufficiently large finite simple group $G$ . Part (ii) of the Theorem generalises this result (and implies it with $w_1=w_2=w_3=w$ ). Moreover, the authors conjectured that the same is true for just 2 group words! Part (i) of the Theorem proves this conjecture for the alternating groups.