Sylow subgroups of quotient groups [closed]

Question

Could somebody explain in detail the steps required to solve this problem?

Let $p$ be a prime dividing $o(G)$. Show that every Sylow $p$-subgroup of $G/K$ is of the form $PK/K$, where $P$ is a Sylow $p$-subgroup of $G$.

Answer 1

I'll sketch some steps for you:

(i) Suppose $|K| = p^a m$ and $|G| = p^{a + b} mn$, where $m$ and $n$ are not divisible by $p$. Then $|G / K| = p^b n$. What is the order of a Sylow $p$-subgroup of $G / K$?

(ii) Show that every subgroup of $G / K$ is of the form $H/K$, where $H$ is some subgroup of $G$ containing $K$. Deduce that any Sylow $p$-subgroup of $G / K$ is of the form $H/K$ for some subgroup $H$ of order $p^{a + b}m$.

(iii) By Sylow's theorems, this $H$ contains a Sylow $p$-subgroup $P$. What is the order of $P$? Convince yourself that $P$ is also a Sylow $p$-subgroup of $G$.

(iv) If you can now show that $ PK = H$, then you are done. Since you already know that $P$ and $K$ are both subgroups of $H$, it is enough (why?) to show that $|PK| = |H|$, i.e. that $|PK| = p^{a+b}m$. But $|P| = p^{a + b}$ and $|K|=p^a m$, and $m$ is not divisible by $p$. Now apply Lagrange's theorem...