Intersection of all $p$-Sylow subgroups is normal

Question

Let $G$ be a finite group, $p$ a prime number that divides $|G|$ and $O_p(G)=\bigcap_{P \in Syl_p(G)}P$. Prove that

1) $O_p(G) \lhd G$

2) $O_p(G)$ is maximal among the normal $p$-subgroups of $G$.

3) $O_p(G/O_p(G))=1$

I think I could do 2) using the following proposition which I will just enunciate:

Proposition

Let $P$ be a $p$-Sylow and $H \subset G$ a $p$-subgroup. Then there is $x \in G$ such that $xHx^{-1} \subset P$.

Solution to 2):

Suppose $O_p(G) \subset H$ for some $H \lhd G$ p-subgroup. By the proposition we have that for each $P$ p-Sylow group, there is $x: H=xHx^{-1} \subset P$. But then $H \in \bigcap_{P \in Syl_p(G)} P=O_p(G)$, so $O_p(G)=H$, it follows $O_p(G)$ is maximal.

I am pretty lost in 1) and 3), I would appreciate suggestions or hints rather than complete answers. Thanks in advance.

Answer 1

For (1), remark that if $\sigma\in \text{Aut}(G)$, then $\sigma$ takes $p$-Sylow subgroups to $p$-Sylow subgroups. It follows that $\sigma$ takes the intersection of all $p$-Sylow subgroups to itself, i.e. $\sigma(O_p(G)) = O_p(G)$. Thus $O_p(G)$ is a characteristic subgroup. In particular, it is a normal subgroup.

For (3), use (2); show that $G/O_p(G)$ has no nontrivial normal $p$-subgroup. (Suppose that $H/O_p(G)$ is a normal $p$-subgroup of $G/O_p(G)$; then show that $H$ is a normal $p$-subgroup of $G$ containing $O_p(G)$...)

Answer 2

Nice facts to add: there is a famous theorem of Jerald Brodkey (see for instance the book of I.M. Isaacs, Finite Group Theory, 1.37), saying that if the Sylow $p$-subgroups are abelian, then the $O_p(G)$ can be realized as the intersection of a certain pair of different Sylow $p$-subgroups. If $G$ has exactly $p+1$ Sylow $p$-subgroups, then $O_p(G)$ can be realized as the intersection of any pair of different Sylow $p$-subgroups!