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.

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