I'm supervising an undergraduate research project. Among other things, I've got the student to look at [this paper of Gene Kopp and John Wiltshire-Gordon](http://arxiv.org/abs/1102.4353). This question arose from a missing complex conjugate in something the student wrote.


Let $g$ be an element of a finite group $G$, and $w$ a word in $n$ variables. If you evaluate the word on all $n$-tuples of elements of $G$, does it give $g$ and $g^{-1}$ the same number of times?


I thought the answer must be "no", but found it frustratingly difficult to come up with an example. After tapping local knowledge it seems that I was right. There's a [recent paper of Alexander Lubotzky](http://arxiv.org/abs/1211.6575) that proves that if $G$ is a finite simple group then the only restriction on a subset $A\subseteq G$ for it to be the image of the word map for some word in 2 variables is that it contains the identity and is fixed by $\operatorname{Aut}(G)$. Since there are finite simple groups (e.g., the Mathieu group $M\_{11}$) with elements that are not sent to their inverses by any automorphism, this answers the question. 


However, my real question is whether there's a relatively simple example? 


Lubotzky's paper doesn't give an explicit word, although it does show that, for $M\_{11}$, there's a word that works with length at most about $1.7\times 10^{244552995}$. 


Presumably one can do a bit better than that?


There are obvious restrictions on $g\in G$ and $w$ that rule out *really* small examples. There can't be any automorphism of $G$ sending $g$ to $g^{-1}$ or any automorphism of the free group $F\_n$ sending $w$ to $w^{-1}$. 



Word evaluating to a group element and its inverse with different frequency
Yes, one can do much better than $1.7 \times 10^{244552995}$
(not surprisingly, because we're asking less than Lubotzky:
one of the two counts must be less than the other, 
but not necessarily zero).
In fact a word of length $10$ suffices.


I tried $G = M\_{11}$ and $g$ an element of order $11$, and took $n=2$,
which makes exhaustive computation easily feasible (the first variable 
can be assumed to lie in one of the $10$ conjugacy classes so 
there's only $10 \, |M\_{11}| = 79200$
group elements to compute given $w$).
None of the words $w(x,y) = x^a y^b x^c y^d$ seems to work, 
but several of the form $w(x,y) = x^a y^b x^c y^d x^e$ solve the problem. 
The first one (in lexicographic order) with all exponents at most $3$ is
$(a,b,c,d,e) = (1,2,1,3,3)$, i.e. $w(x,y) = x y^2 x y^3 x^3$,
for which $w(x,y) = g$ has $7491$ solutions 
but $w(x,y) = g^{-1}$ has only $7458$.


