Show that there exists a Sylow p-group P of G such that $P_1 = P\cap H$.

Question

Here is the statement needs to prove,

Let $G$ be a finite group, $H$ is the subgroup of $G$. Let $P_1$ be a Sylow $p$-group of $H$. Show that there exists a Sylow $p$-group $P$ of $G$ such that $P_1 = P\cap H$.

I think we should start with $H$ act on $G/P$, then the stabilizer of $G/P = P\cap H$. But I don't know how to prove stabilizer of $G/P$ is equal to $P_1$.

Answer 1

Suppose $|H| = p^am$ and $|G| = p^bmn$, where $b \geq a$ and $gcd(p,m)=gcd(p,n) = 1$. Then $|P_1| = p^a$.

One of Sylow's theorems says that there exists a Sylow subgroup $P \subset G$ such that $P_1 \subset P$.

But $|P| = p^b$, so by Lagrange's theorem, $|P \cap H| = p^c$ where $c \leq a$. Moreover, $P_1 \subset P \cap H$, so in fact, $c = a$ and $P \cap H = P_1$.

Answer 2

You can use the reference which shows that every $p$-group is contained in a Sylow $p$-subgroup as follows: There exists $P$ a Sylow $p$-subgroup of $G$ such that $P_1\subset P$. You have $P\cap P_1=P_1$.

Show a certain group is contained in a Sylow p-group.