Group $G$ of order $55$ acts on a set $M$ of cardinality $39$. Prove, that there is an element $x \in M$, such that $\forall g \in G$ $gx=x$.
I tried to apply some combinations of Burnside's lemma or Orbit-stabilizer theorem, to compute sizes of stabilizer subgroups but didn't achieve anything.
Lengths of orbits under $G$ can be $1,5,11,55$. As $M$ cannot have an orbit of length $55$, assume that there are only $a$ orbits of lengths $5$ and $b$ orbits of length $11$. Then $5a+11b=39$. This has no solution in $a,b\ge0$.