## In a finite group, if the number of Sylow subgroups is not 1 mod p² then there exist Sylow subgroups which intersect maximally

Let $G$ be a finite group. Use the method of proof in Sylow’s Theorem to show that if $n_p \neq 1$ mod $p^2$, then there exist distinct Sylow $p$-subgroups $P$ and $Q$ such that $[P : P \cap Q] = [Q : P \cap Q] = p$.

Let $\mathcal{S} = \{ P_1, P_2, \ldots, P_r \}$ be the Sylow $p$-subgroups of $G$.

Now let $Q$ be any $p$-subgroup of $G$. Note that $Q$ acts on $\mathcal{S}$ by conjugation; we can then say $\mathcal{S} = \bigcup_{P_i \in \mathcal{S}} Q \cdot P_i$. Relabel the elements of $\mathcal{S}$ so that $P_i$ for $1 \leq i \leq s$ are a set of orbit representatives of the action of $Q$. Note that $|Q \cdot P_i| = [Q : N_Q(P_i)]$ by the Orbit-Stabilizer Theorem, so that $|Q \cdot P_i| = [Q : N_G(P_i) \cap Q] = [Q : P_i \cap Q]$ by Lemma 19 in the text. Thus $n_p = \sum_{i=1}^s [Q : P_i \cap Q]$, where $P_i$ ranges over a set of orbit representatives of the action of $Q$ on $\mathcal{S}$.

We may let $Q$ itself be a Sylow $p$-subgroup, and (without loss of generality and with a relabeling if needed) choose $P_1 = Q$. Then $N_p = 1 + \sum_{i=2}^s [Q : P_i \cap Q]$. Now because $P_i \cap Q$ is a $p$-subgroup, $[Q : P_i \cap Q] = p^{k_i}$ for some $k_i$. Modulo $p^2$, then, $[Q : P_i \cap Q] \in \{0,p\}$. If $[Q : P_i \cap Q] = 0$ (mod $p^2$) for all $P_i$, then we have $n_p \equiv 1 \mod p^2$, a contradiction. Thus there exists a Sylow $p$-subgroup $P$ such that $[Q : P \cap Q] = p$.

Finally, note that $[P : P \cap Q] = |P|/|P \cap Q| = |Q|/|P \cap Q| = [Q : P \cap Q] = p$, by Lagrange and since $|P| = |Q|$.