A subgroup $H \le G$ is called permutable, if $UH = HU$ for every subgroup $U \le G$. Show that if $H$ is a permutable $\pi$-group, then its normal closure $H^G$ is a $\pi$-group. Deduce that if $H$ is a $p$-group, then $H^G$ is nilpotent.
Does anyone has a proof of this fact?
I will assume that $H$ is finite, because otherwise the final statement is false.
If two finite $\pi$-groups permute, then their product is also a $\pi$-group. So, you can prove by induction on $n$ that the product of any $n$ conjugates of $H$ is a $\pi$-group, for any $n \ge 0$.
Now, $H^G$ is generated by the conjugates of $H$, so any $g \in H^G$ is a product $h_1^{g_1} \cdots h_n^{g_n}$ of $n$ conjugates of elements of $H$ for some $n \ge 0$, and hence $g$ lies in the $\pi$-group $H^{g_1}H^{g_2} \cdots H^{h_n}$, so $g$ is a $\pi$-element, and hence $H^G$ is a $\pi$-group.