Group of order $p^{\alpha}q$ is not simple.

$$|G|=p^{\alpha}q$$, 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 $$D and $$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.

• Where did you find the proof above? I find it not easy to make sense of it, so I want to learn the previous content of the material. Mar 15, 2022 at 10:13

$$P_1\cap P_3 \geq P_1\cap N_G(D)=N_{P_1}(D)$$
Since the Sylow $$p$$-subgroups are all pairwise conjugate (Sylow II), their intersection is the normal core in $$G$$ of any of them. So, once you know this intersection is not trivial, you get a nontrivial normal subgroup of $$G$$, which is then nonsimple.