Let $G$ be finite with $G > 1$ and suppose $P \subseteq Aut(G)$ is a $p$-subgroup. Show that there exists some nontrivial Sylow $q$-subgroup $Q$ of $G$ (for some prime $q$) such that $\sigma(Q)=Q$, $\forall$ $\sigma$ $\in P$.
I don't know how to attack the problem. Thanks in advance for your help!
Hints:
The group $\;P\;$ acts on the set $\;Syl_q\;$of all the Sylow $\;q$-subgroups of $\;G\;$ , for each prime $\;q\;$ dividing $\;|G|\;$, by $\;\;\;\sigma\cdot Q:=\sigma(Q)\;,\;\;\forall\,Q\in S_q\;$ . By the orbit-stabilizer theorem, we get that
$$|\mathcal O(Q)|=[P:P_Q]\;,\;\;P_Q=\text{Stab}(Q):=\{\pi\in P\;;\;\;\pi(Q)=Q\;\}$$
If $\;p\mid |G|\;$ then $\;|Syl_p|=1\pmod p\;$ and we're done as it must be $\;|\mathcal O(Q)|=1\;$ for some $\;Q\in Syl_p\;\;\;$ (why?)
Otherwise (i.e., there exists a prime $\;q\neq p\;$ dividing the group's order), using $\;Syl_q^P:=\{ Q\in Syl_q\;:\;\;\sigma(Q)=Q\;,\;\;\forall\sigma\in P\}\;$ , the fixed point theorem gives us
$$|Syl_q^P|=|Syl_q|\pmod p$$
Finish the proof now.