$P$ is a Sylow subgroup of $G$ then $P$ is normal in $G$.

Question

Let $G$ be a finite group with subgroups $H$ and $P$ and if $H$ is normal in $G$ and $P$ is normal in $H$ and $P$ is a Sylow subgroup of $G$ then $P$ is normal in $G$.

Is the statement true, I heard it from a friend of mine....but helpless to do the proof.

Answer 1

If $P$ is a $p$-Sylow subgroup of $G$, then it is also a $p$-Sylow subgroup of $H$ since it is still a $p$-group and certainly maximal.

For any $g\in G$, you have $gPg^{-1}\subseteq gHg^{-1}=H$ (using normality of $H$), so $gPg^{-1}$ is again a $p$-Sylow-Subgroup of $H$. By the second Sylow Theorem, all $p$-Sylow subgroups (of $H$) are conjugate, so there must be an $h\in H$ with $gPg^{-1}=hPh^{-1}=P$, where we use that $P$ is normal in $H$.

Answer 2

Let $|G|=p^{\alpha}m$, where $p\nmid m$. $|P|=p^{\alpha}$. Since $P\leq H$, $|H|=p^{\alpha}k$ for some $k\in \mathbb{Z}$. Since $H\leq G$, $k|m$. So $p\nmid k$. Hence $P\in Syl_{p}(H)$. Since $P\unlhd H$, $P$ char $H$. Since $P$ char $H \unlhd G$, $P\unlhd G$.