MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 5 down vote accepted

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.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.