# 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?

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You could ask for words up a certain length. This is a well-studied question. For example it is elementary to see that for any word of length $n$, there are permutations $\sigma,\tau \in S_{n+1}$, such that $w(\sigma,\tau)\neq 1_{n+1}$. – Andreas Thom Apr 14 '14 at 12:32
If you let $\mathfrak{V}$ be the variety generated by $S$, then any word in $\mathfrak{V}(F)$ would fail the property; and as Derek Holt's answer shows, the subgroup $\mathfrak{V}(F)$ does not consist only of power words. – Arturo Magidin Apr 14 '14 at 16:03

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.

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

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Yes that is a trivial counterexample. How can one avoid this things? what should I need to assume about w to bypass this? – Pablo Apr 14 '14 at 11:14
At least, you should add that the word does not contain $n$th power of a word, where $n$ is the period of $S$; otherwise you may have words like $a^kb^na^{n-k}$ etc. – Ilya Bogdanov Apr 14 '14 at 11:29
More generally, take anything in the normal subgroup of $F$ generated by $n$th powers. – Eric Wofsey Apr 15 '14 at 6:17