## Sylow subgroup intersection of a certain index + F1 Trespasser

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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 $Q_i$.

Thus, $|O(Q_i)|=|P|/|Q_i\cap P|\geq q^a$, i.e. $q^a\mid |O(Q_i)|$. $|O(P)|=1$.

The orbits form a partition of $\Omega$, thus $|\Omega|=1+\sum{|O(Q_i)|}$, where the sum runs over all orbits other than $O(P)$.

Thus, $n_q\equiv 1\pmod {q^a}$.

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