# Finite groups: equations with many solutions

Let $\omega$ be a word in the free group on generators $x_1,x_2,\ldots,x_n,g_1,g_2,\ldots,g_k$, where $n>0$ and $k\geq 0$. For any finite group $G$ and elements $g_1,g_2,\ldots g_k$ in $G$ we obtain an equation: $$\omega(x_1,x_2,\ldots,x_n,g_1,g_2,\ldots,g_k)=e$$ Let $S=S_{\omega,g_1,g_2,\ldots,g_k}$ be the set of its solutions in $G^n$.

Question: Is it true that for every $\omega$ there exists a constant $c<1$ such that for every finite group $G$ and its elements $g_1,g_2,\ldots,g_k$ if $|S|>c|G|^n$ then $S=G^n$?

Comments: There exist many papers on problems related to the number of solutions of such equations (look here for an example) but I couldn't find the above anywhere...

Some very easy cases: for $\omega(x,g)=xgx^{-1}g^{-1}$ we have $c={1\over 2}$, for $\omega(x_1,x_2)=x_1x_2x_1^{-1}x_2^{-1}$ we may take $c={3\over 4}$.

Why do I ask: if the answer is yes then for some very nice subsets of a profinite group positive Haar measure would imply nonempty interior.

• For $w(x_1,x_2)=x_1x_2x_1^{-1}x_2^{-1}$, we have $c=\frac{5}{8}$. Of course, any greater number also works. Commented Feb 3, 2016 at 6:03

In the case $k=0$ this is a conjecture of Dixon. See http://people.math.carleton.ca/~jdixon/Prgrpth.pdf, Section 1.2. I don't know of any reason it should become obviously false for $k>0$, so I guess it's open.