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The following is a classic result: Let $G$ be a solvable finite group. Assume that $G$ has an abelian Hall $\omega-$subgroup $H$ and that $O_{\omega'}(G)=1$. Then $H\triangleleft G$.

Is the result true if we replace abelian by nilpotent ?


The proof of the classic result can go as as follows:

We argue by induction on $|G|$. If $|G|=1$, there is nothing to prove. We assume that $H<G$ otherwise the result is trivial.

We show that for every proper normal subgroup $K$ of $G, H\cap K\triangleleft G$. $O_{\omega'}(K)=K\cap O_{\omega'}(G)=1$. Moreover, $H\cap K$ is an abelian Hall $\omega-$subgroup of $K$. Therefore, by the induction hypothesis applied to $K$, $H\cap K \triangleleft K$. Now $H\cap K \sqsubset K\triangleleft G$, and so $H\cap K\triangleleft G$.

Since $O_{\omega'}(G)=1$, it follows by that $O_{\omega}(G)>1$ ($G$ is $\omega-$separable). Consider $C=C_G(O_{\omega}(G))$. We show that $C<G$. It can be done

-A) By transfer: Deny it. Then $O_{\omega}(G)\le Z(G)$. Since $H$ is abelian, by a transfer result, $$H\cap G'\cap Z(G)=1$$ Therefore, $H\cap G'\cap O_{\omega}(G)=1$. Now $H\cap G'$ is a normal $\omega-$subgroup of $G$ : it is obviously an $\omega-$group and since $G'<G$ (otherwise $G$ would be perfect and thus not solvable), it follows by 1) that $H\cap G'\triangleleft G$. Therefore, $H\cap G'\le O_{\omega}(G)$ and so $H\cap G'=1$. But then, it means that $H$ has a normal complement in $G$. This normal complement must be $O^{\omega}(G)=O_{\omega'}(G)=1$, leading to a contradiction as $H<G$.

-B) By the Hall-Higman lemma, $C\le O_{\omega}(G)$. If $O_{\omega}(G)=G$, then $H=G$, contradiction. Therefore $O_{\omega}(G)<G$ and the result follows.

Since $C$ is a proper normal subgroup of $G$, it follows by 1) that $H\cap C\triangleleft G$. Now we show that $H\cap C=H$, which allows to conclude. $O_{\omega}(G)$ is a $\omega-$subgroup of $G$ and $G$ is solvable therefore it is contained in a Hall $\omega-$subgroup of $G$ by Hall D. As $O_{\omega}(G)\triangleleft G$, it is contained in $H$ by Hall C. Since $H$ is abelian, $[H,O_{\omega}(G)]=1$ i.e. $H\le C$ and so $H\cap C=H$.

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No it is not true if you replace abelian by nilpotent. Since $p$-groups are nilpotent, we can take $H$ to be any Sylow $p$-subgroup of $G$. A counterexample is $S_4$ with $p=2$.

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