Q:

Simple groups and words

Let S be a finite simple nonabelian group, w a word in a finite number of variables which is not a power of another word. Must there be a substitution of elements of S in w such that the resulting element is not 1?
Equivalently, let S be a finite simple group and F a free group on a finite number of variables. Let w be an element of F which is not a power of another element in F. Is there a homomorphism from F to S for which w is not in the kernel?

A:

Let $a$ and $b$ be in a free generating set of $F$ and $w = a^{|S|}b^{|S|}$.

A:

The answer is 'no' if you fix $S$ and consider all (primitive) words $w$ (see Derek Holt's answer).  However, if you fix the word $w$, then $w$ takes a non-trivial value on $S$ (equivalently, $S$ is generated by values of the word $w$) for all but finitely many non-abelian finite simple groups $S$.  This is due to the fact that any infinite family of non-abelian finite simple groups generates the variety of all groups, as proved here:
G. A. Jones, Varieties and simple groups, J. Austral. Math. Soc. 17 (1974), 163–173.
Much stronger results have since been proven.  For instance, there is the result of Shalev that given a non-trivial word $w$, then in all but finitely many non-abelian finite simple groups, every element is a product of three $w$-values:
A. Shalev, "Word maps, conjugacy classes, and a non-commutative Waring-type theorem", Annals of Math. 170 (2009), 1383-1416.

