## If any Sylow p-subgroup intersects all others trivially, then all Sylow p-subgroups intersect trivially

Let $G$ be a finite group and $p$ a prime. Suppose that for some $P \in \mathsf{Syl}_p(G)$, we have $P \cap R = 1$ for all $R \in \mathsf{Syl}_p(G)$ with $R \neq P$. Prove that $P_1 \cap P_2 = 1$ for all $P_1,P_2 \in \mathsf{Syl}_p(G)$. Deduce that in this case, the number of nonidentity elements of $p$-power order in $G$ is precisely $(|P|-1)[G:N_G(P)]$.

Let $P_1$ and $P_2$ be distinct Sylow $p$-subgroups. Now by Sylow’s Theorem, $P_1$ is conjugate to $P$ in $G$; say $xP_1x^{-1} = P$. Now if we also have $xP_2x^{-1} = P$, then $P_1 = x^{-1}Px = P_2$, a contradiction. Note then that $x(P_1 \cap P_2)x^{-1} = xP_1x^{-1} \cap xP_2 x^{-1}$ $= P \cap Q$, where $Q$ is some Sylow $p$-subgroup of $G$ distinct from $P$. Thus $x(P_1 \cap P_2)x^{-1} = 1$, and we have $P_1 \cap P_2 = 1$.

Now every nonidentity element of $p$-power order is in some Sylow $p$-subgroup of $G$, an no such element is in more than one. The number of nonidentity elements in a Sylow $p$-subgroup is $|P|-1$ and the number of Sylow $p$-subgroups is $n_p(G) = [G:N_G(P)]$, and the final conclusion follows.