Exercise 3.3.9 (The intersection of any Sylow $p$-subgroup with a normal subgroup $N$ is a Sylow $p$-subgroup of $N$)

Question

Let $p$ be a prime and let $G$ be a group of order $p^a m$, where $p$ does not divide $m$. Assume that $P$ is a subgroup of $G$ of order $p^a$ and $N$ is a normal subgroup of $G$ of order $p^b n$, where $p$ does not divide $n$. Prove that $|P \cap N| = p^b$ and $|PN/N| = p^{a-b}$. (The subgroup $P$ of $G$ is called a {\bf Sylow $p$-subgroup of $G$}. This exercise shows that the intersection of any Sylow $p$-subgroup with a normal subgroup $N$ is a Sylow $p$-subgroup of $N$.)

Richard Ganaye · Accepted Answer

\begin{center}
\begin{tikzpicture}
    
ode (n0) at (3,0) {$P \cap N$};
    
ode (n1) at (1.5,1.5) {$P$};
    
ode (n4) at (4.5,1) {$N$};
    
ode (n5) at (3,2.5) {$PN$};
    
ode (n6) at (3, 3.5) {$G$};
     \draw (n0) edge (n1)  edge (n4)  ;
     \draw (n5) edge (n1)  edge (n4) edge (n6) ;
   \end{tikzpicture}
\end{center}
\begin{proof} 
Put $k = |P \cap N$. Since $P \cap N \leq P$ and $ P \cap N \leq N$, we obtain by Lagrange's Theorem that $k \mid p^a$ and $k \mid p^b n$. Since $k \mid p^a$, then $k = p^\beta$ for some integer $\beta \leq a$. Then $k = p^\beta \mid p^b n$, where $p$ is relatively prime with $n$, thus $p^\beta \mid p^b$, so 
\begin{align}
 |P \cap N | = p^\beta, \qquad \beta \leq b.
 \end{align}

Since $N \unlhd G$, $PN$ is a subgroup of $G$, and the Second Isomorphism Theorem shows that $P\cap N \unlhd P$ and
$$PN/N \simeq P/P\cap N.$$
Since $PN \leq N$, then $PN/N \leq G/N$, so
$$|P : P \cap N| = |PN : N | 	ext{ divides } |G/N|.$$
This gives $(p^a/p^\beta) \mid (p^a m/ p^b n)$ (where $n \mid m$ because $n \mid p^a m$ and $n\wedge p = 1$). Therefore
$$p^{a-\beta} \mid p^{a-b}\,  \frac{m}{n}.$$
Since $p \wedge m = 1$, then $p^{a-\beta} \wedge \frac{m}{n} = 1$, thus
$$p^{a-\beta} \mid p^{a-b}.$$
Therefore $a - \beta \leq a -b$, so $b \leq \beta$. If we compare with (1), we obtain
$$\beta = b.$$
This gives
\begin{align*}
|P \cap N| &= p^b,\
|PN/N| &= |P/P\cap N| = p^{a-b}.
\end{align*}
(Since $|N| = p^b n,\ n \wedge p = 1$, $P \cap N$ is a $p$-Sylow of $N$.)
\end{proof}

Exercise 3.3.9 (The intersection of any Sylow $p$-subgroup with a normal subgroup $N$ is a Sylow $p$-subgroup of $N$)

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Exercise 3.3.9 (The intersection of any Sylow $p$-subgroup with a normal subgroup $N$ is a Sylow $p$-subgroup of $N$)

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