Sylow subgroup intersection of a certain index + F1 Trespasser

Today, I read the news online, the latest news is that a man was strolling along the F1 track while the race was ongoing. Really unbelievable.

Also, recently our recommended books from Amazon for GAT/DSA preparation have been very popular with parents seeking preparation. Do check it out if your child is going for DSA soon.

Back to our topic on Sylow theory…

Let $G$ be a finite group, where $q$ is a prime divisor of $G$ . Suppose that whenever $Q_1$ and $Q_2$ are two distinct Sylow q-subgroups of $G$ , $Q_1\cap Q_2$ is a subgroup of $Q_1$ of index at least $q^a$ . Prove that the number $n_q$ of Sylow q-subgroups of G satisfies $n_q\equiv 1\pmod {q^a}$ .

Proof: Let $\Omega=\{Q_1,Q_2\dots,Q_n\}$ be the set of all Sylow q-subgroups of G. Fix $P=Q_k\in \Omega$ . Consider the group action of P acting on $\Omega$ by conjugation.

$\phi:P\times\Omega\to\Omega$ , $\phi_x(Q_i)=xQ_i x^{-1}$

By Orbit-Stabilizer Theorem, $|O(Q_i)|=|P|/|N_p(Q_i)|$ .

We claim that $N_p(Q_i)=Q_i\cap P$ , since any element x outside of $Q_i$ cannot normalise $Q_i$ , since otherwise if $x \neq Q_i$ , $xQx^{-1}=Q_i$ , then $\langle Q_i, x\rangle$ will be a larger q-subgroup of G than $Q_i$ .