$|G|=p^{\alpha}q$, where $p,q$ are distinct primes, $\alpha \geq 1$. Show $G$ is not simple.

I am trying to follow a proof and I understand all of it except one part which is blocking me.

The proof begins by assuming $G$ is simple. First it shows that the $p$ Sylow subgroups cannot intersect trivially (counting argument).

Then take $P_1 \neq P_2$ to be two Sylow subgroups such that their intersection, $D=P_1\cap P_2$ is maximal. Then $D < P_1,P_2$ so $D<N_{P_1}(D)$ and $D<N_{P_2}(D)$. Consider $N_G(D)$. If $N_G(D)$ is a $p$ subgroup then by Sylow it is contained inside $P_3$, which is a $p$ Sylow group.

$\textbf{This next line I cannot understand, this is what I want explained.}$

$P_3 \cap P_1 \geq N_{P_1}(D) > D \implies P_3 = P_1$ (*).

I don't understand why the intersection contains the normalizer. I will include the rest of the proof for anyone who looks here in the future.

And similarly $P_3 = P_2$, so $P_1 = P_2$, so $N_G(D)$ is not a $p$ group. Thus a $q$ Sylow group lies in $N_G(D)$. Then $P_1N_G(D) = G$ and if we pick $g\in G$, then $g=hx$, $h\in P, x\in N_G(D)$ and then $P_1^g = P_1^{hx}=P_1^x \geq D^x = D $, and thus $D$ lies in every $p$ Sylow subgroup of $G \implies 1 < D \leq \cap_{g\in G}P_1^g \triangleleft G$, which is a contradiction.


$P_1\cap P_3 \geq P_1\cap N_G(D)=N_{P_1}(D)$

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  • $\begingroup$ I can't believe it was that simple, I spent an embarrassingly long time trying to figure that out. $\endgroup$ – pureundergrad Jan 11 '19 at 18:00

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