Show that the intersection is a pp -Sylow subgroup

[mathmari]

Hey!

I want to show that if $S\in \text{Syl}_p(G)$ and $N\trianglelefteq G$, then $N\cap S\in \text{Syl}_p(N)$. Could you give me some hints how we could show that? (Wondering)

Do we maybe use Frattini's Argument? (Wondering)
From that we have that since $N\trianglelefteq G$ and $S\in \text{Syl}_p(G)$, $G=NN_G(P)=N_G(P)N$, right? (Wondering)
But does this help us? (Wondering)

 

[mathmari]

I have done the following:

Since $S\in \text{Syl}_p(G)$, i.e., $S\leq G$, and $N\trianglelefteq G$ we have that $NS\leq G$.

We have that $|NS|=\frac{|N||S|}{|N\cap S|}\Rightarrow \frac{|NS|}{|S|}=\frac{|N|}{|N\cap S|}\Rightarrow [NS:S]=[N:N\cap S]$

Since $S\in \text{Syl}_p(G)$ we have that $[G:S]$ is coprime with $p$.
We have that $[G:S]=[G:NS][NS:S]$.
Since $[NS:S]\mid [G:S]$, $[N:N\cap S]=[NS:S]$ is coprime with $p$.
Since $N\cap S\leq S$, from Lagrange's theorem we have that $|N\cap S|\mid |S|$.
Since $S\in \text{Syl}_p(G)$, if $|G|=p^na$, with $p\not\mid a$, then $|S|=p^n$.
Then $|N\cap S|=p^m$, with $0<m\leq n$.
Since $|N\cap S|=p^m$ and $[N:N\cap S]$ is coprime with $p$, we have that $N\cap P\in \text{Syl}_p(N)$.

Is everything correct? (Wondering)

 

[Deveno]

I have done the following:

Since $S\in \text{Syl}_p(G)$, i.e., $S\leq G$, and $N\trianglelefteq G$ we have that $NS\leq G$.

We have that $|NS|=\frac{|N||S|}{|N\cap S|}\Rightarrow \frac{|NS|}{|S|}=\frac{|N|}{|N\cap S|}\Rightarrow [NS:S]=[N:N\cap S]$

Since $S\in \text{Syl}_p(G)$ we have that $[G:S]$ is coprime with $p$.
We have that $[G:S]=[G:NS][NS:S]$.
Since $[NS:S]\mid [G:S]$, $[N:N\cap S]=[NS:S]$ is coprime with $p$.
Since $N\cap S\leq S$, from Lagrange's theorem we have that $|N\cap S|\mid |S|$.
Since $S\in \text{Syl}_p(G)$, if $|G|=p^na$, with $p\not\mid a$, then $|S|=p^n$.
Then $|N\cap S|=p^m$, with $0<m\leq n$.
Since $|N\cap S|=p^m$ and $[N:N\cap S]$ is coprime with $p$, we have that $N\cap P\in \text{Syl}_p(N)$.

Is everything correct? (Wondering)

It looks like it to me.

 

[mathmari]

It looks like it to me.

Ah ok... Thank you! (Smile)