# Constructing a group which has fixed-point free element and element with exactly $n$ fixed points

Given some $n \ge 1$, how could we construct groups $G$ such that there exists some non-trivial element fixing no points and some non-trivial element fixing exactly $n$ points.

If given a transitive permutation group $G$ acting on $\Omega$, and $\alpha_1,\ldots, \alpha_n \in G$, I guess somehow the point stabilizer $G_{\alpha_1,\ldots, \alpha_n} = \bigcap_{i=1}^n G_{\alpha_i}$ acting on $\Omega$ would be a good canditate, but this construction is not general enough.

• You could take $G=S_k$ for any $k \ge n+2$. But there are lots of examples. – Derek Holt Oct 12 '15 at 14:10

As Derek Holt points out, $S_{n+2}$ with the natural action works. But there are smaller groups as well, even under the constraint that the action is transitive; for example, take $(\mathbf{Z}/n\mathbf{Z})\times S_3$ acting diagonally on $[n]\times[3]$; $(e,(12))$ has $n$ fixed points and $(r,*)$ has no fixed points if $r\ne e$.