Let $G$ be a finite group. I want to prove that
If $N$ is a subnormal $\pi$-subgroup of $G$, then $N\le O_{\pi}(G)$.
I first tried the case where $\pi=\{p\}$, $p$ a prime, but I got stuck.
Here are my attempts:
Since $N\trianglelefteq\trianglelefteq G$, there exist $G_1,G_2,\cdots,G_{d-1},G_d$ such that
$$N=G_1\trianglelefteq G_2\trianglelefteq\cdots\trianglelefteq G_{d-1}\trianglelefteq G_d=G.$$
$N$ is normal in $G_2$, so $N\le O_p(G_2)$. The Sylow $p$-subgroups of $G_2$, as $p$-subgroups of $G$, are contained in Sylow $p$-subgroups of $G$. Then
$$N\le O_p(G_2)=\bigcap_{P\in {\rm Syl}_pG_2} P\le\bigcap_{P\in {\rm Syl}_pG} P=O_p(G). $$
But I made a mistake. I can only show that the Sylow $p$-subgroups of $G_2$ are contained in some Sylow $p$-subgroups of $G$, but I didn’t show they are contained in each Sylow $p$-subgroup of $G$. So we can’t conclude that $N$ is contained in $O_p(G)$, which is the intersection of all the Sylow $p$-subgroups of $G$.
My question is: how to prove it in the case where $\pi=\{p\}$ and how to prove it in general?
Any help is appreciated. Thanks!
Best Answer
Observe that $O_p(G_2) \text{ char } G_2 \unlhd G_3$, so $O_p(G_2) \unlhd G_3$. It follows that $O_p(G_2) \subseteq O_p(G_3)$. So with induction, $O_p(G_2) \subseteq O_p(G)$. Hence $N \subseteq O_p(G)$. This reasoning generalizes to $O_\pi(G)$.
Let me make this more precise. What we will use and what I have been using above is the fact that if $X$ is a characteristic subgroup of $Y$, and $Y \unlhd Z$ then $X \unlhd Z$, Here $X$ is characteristic in $Y$ if every automorphism of $Y$ maps $X$ onto $X$ and we write $X$ char $Y$.
Now recall that $O_\pi(G)$ is the unique largest normal $\pi$-subgroup of $G$ (or, differently put, it is the intersection of all Hall $\pi$-subgroups). One can easily show that $O_\pi(G)$ is characteristic in $G$.
Let $S$ be a subnormal $\pi$-subgroup of $G$, say $S=H_0 \lhd H_1 \lhd \cdots \lhd H_r=G$. Since $S$ is normal, $S \subseteq O_\pi(H_1)$. Observe that $O_\pi(H_1) \text{ char } H_1 \lhd H_2$, so $O_\pi(H_1) \lhd H_2$ and this yields $O_\pi(H_1) \subseteq O_\pi(H_2)$. In its turns, $O_\pi(H_2) \text{ char } H_2 \lhd H_3$, so $O_\pi(H_2) \lhd H_3$ and this yields $O_\pi(H_2) \subseteq O_\pi(H_3)$. Now work up your way till $H_r=G$ is reached and we conclude $S \subseteq O_\pi(H_1) \subseteq O_\pi(H_2) \subseteq \cdots \subseteq O_\pi(G)$.
As a corollary we can also conclude that the subgroup generated by two subnormal $\pi$-subgroups of $G$ is again a $\pi$-subgroup.