Proof: Note that since $p$-groups are nilpotent, $P∩Q < N_P(P∩Q)$ and $P∩Q < N_Q(P∩Q)$, so that $p \cdot |P∩Q|$ divides $|N_G(P∩Q)|$. Prove that $N_P(P∩Q)$ and $N_Q(P∩Q)$ are distinct Sylow $p$-subgroups of $N_G(P∩Q)$. If $N_P(P∩Q)$ is not a Sylow subgroup, then place it in one, and call this group $P^*$. This is a $p$-group in $G$, so place $P^*$ in a Sylow $p$-subgroup $P^{**}$. Note that if $P^{**}=P$, then since $P^* ≤ N_G(P∩Q)$ we have $P^*∩P = P^* ≤ N_P(P∩Q)$, a contradiction. Therefore $P$ and $P^{**}$ are distinct Sylow $p$-subgroups of $G$ and $P∩Q < N_P(P∩Q) ≤ P∩P^{**}$, a contradiction by the maximality of $P∩Q$. The case for $N_Q(P∩Q)∈Syl_p(N_G(P∩Q))$ is similar. Now prove that $N_P(P∩Q)≠N_Q(P∩Q)$: Assuming the contrary, we have $P∩Q < N_P(P∩Q) ≤ P∩Q$, a clear impossibility.
Assume $N_G(P∩Q)$ is a $p$-group, and thus place it in a Sylow $p$-subgroup of $G$, calling it $M$. If $M=P$, then $N_Q(P∩Q) ≤ P$, so that $N_Q(P∩Q) ≤ P∩Q$, untenable. Therefore $M≠P$ and $P∩Q < N_P(P∩Q) ≤ M∩P$, another contradiction. Therefore $q$ divides $|N_G(P∩Q)|$.
Finally, let $A,B∈Syl_p(N_G(P∩Q))$ with $A ≠ B$, and prove $A∩B = P∩Q$. ($\supseteq$) Let $P∩Q ≤ C$ for some Sylow $p$-subgroup of $N_G(P∩Q)$ with $gCg^{-1}=A$ for some $g∈N_G(P∩Q)$. We have $g(P∩Q)g^{-1}=P∩Q≤A$. The case is parallel for $P∩Q ≤ B$. ($\subseteq$) $A$ and $B$ are $p$-subgroups of $G$, so let $A≤D$ and $B≤E$ for some $D,E∈Syl_p(G)$. Assume $D=E$: Since $A≠B$ and $A,B≤D$, we have $\langle~A,~B~\rangle$ is a $p$-subgroup (because it is in $D$) properly containing $A$ and $B$, a contradiction (since $\langle~A,~B~\rangle ≤ N_G(P∩Q)$ and $A$ and $B$ are supposed to be Sylow in $N_G(P∩Q)$). So $D≠E$. We have $A∩B≤D∩E$ so that $|A∩B|≤|D∩E|≤|P∩Q|$ and now $A∩B=P∩Q$.$~\square$
No comments:
Post a Comment