Subgroups which contain all $p$-Sylowsubgroups for some fixed prime $p$

Question

Is it true that if some subgroup $H \le G$, $H\ne G$ contains all $p$-Sylowsubgroups for some fixed prime $p$, then $H$ contains some non-trivial normal subgroup of $G$?

Answer 1

Consider the subgroup generated by all elements of order $p$. Any automorphism of $G$ permutes these elements. In particular, this means that the subgroup they generated doesn't change, and thus is characteristic. Each of these elements are also contained in some $p$-Sylow. Since $H$ contains all of the $p$-Sylows and is closed under multiplication, this means that the characteristic subgroup is contained in $H$. Since $H$ is not all of $G$, neither is this subgroup.

Answer 2

Hint: All the conjugates of $H$ also contain all those Sylow $p$-subgroups. The intersection of the conjugates of $H$ is thus...

Answer 3

Yes. $G$ acts by conjugatiopn on the set of $p$-Sylowsubgroups. Hence the smallest subgroup of $G$ that contains all $p$-Sylows is also conjugation-invariant.