TITLE: about fixed points of permutations
QUESTION [5 upvotes]: This is another attempt to make a feasible approximation of this question. Two previous (unsuccessful) attempts are here. 
Let $n\gg 1$ be a fixed number (say, $n=10^{10}$), $k\gg 1$ a natural number. Let $a,b$ be two permutations from $S_k$. Suppose that for every word $w(x,y)$ of length $\le n$, the permutation $w(a,b)$ has a fixed point. Is it true that every permutation in $\langle a, b\rangle$ has a fixed point?

REPLY [3 votes]: Let me have a go at this one. We want to show that for any finite set $W$ of reduced words in $a,b$, we can find a subgroup $G = \langle a,b \rangle$ of some symmetric group $S_k$ such that all words in $W$ have fixed points when evaluated in $G$, but $G$ contains fixed-point-free elements.
Suppose inductively that we can do this for some set $W$, and we want to extend the construction to $W \cup \{w \}$ for some new word $w$. We may suppose that $w$ is cyclically reduced and that $w$ acts fixed-point-freely in $G$.
If $w$ has length $n$, then we can define partial permutations $a,b$ in the obvious way on $n$ points such that $w$ fixes the first of these. For any $p>n$, it is then easy to extend these partial permutations to permutations of a set of size $p$ that generate a transitive subgroup of $S_p$. In fact, by an old theorem of Jordan, if a primitive permutation group of degree $p$ contains a $q$-cycle for some prime $q<p-2$, then it must be the alternating or symmetric group and, by choosing $p$ to be a sufficiently large prime, we can make $b$ contains such a cycle, and thereby ensure that $a,b$ generate $A_p$ or $S_p$.
Do this for some prime $p > k$ to give a group $H \le S_p$, and let $K < S_{k+p}$ be generated by the disjoint actions of $a$ and $b$ in $G$ and in $H$. So all elements of $W \cup \{w\}$ have fixed points in $K$ and it remains to show that $K$ contains fixed-point free elements.
If all relations in $a,b$ that hold in $G$ held in $H$, then $H$ would be a quotient group of $G$, which is impossible, because $|H|$ is divisible by $p$, but $|G|$ isn't. So the permutations generated by the relations of $G$ generate a nontrivial normal subgroup of $H$ which, by the simplicity of $A_p$, must contain $A_p$. So, there exists a relation $v$ of $G$ such that $vw$ acts fixed-point-freely in $H$. Since $w$ acts fixed-point-freely in $G$, $vw$ is a fixed-point-free elemtn of $K$ as required.