Let $G$ be a finite group, $H\leq G$ such that $\forall g\in G$, either $\langle H, H^g \rangle$ is nilpotent or $H H^g = H^g H$. Show that $H \lhd \lhd G$.
I know how to show that if $\langle H,H^g \rangle$ is nilpotent $\forall g\in G$, then $H$ is contained in the Fitting subgroup of $G$, and hence is nilpotent. From here we could conclude that $H \lhd \lhd G$. But we can’t apply that argument here, since there can be some $g\in G$ such that $\langle H,H^g \rangle$ is not nilpotent. I’m a bit stuck.
Best Answer
This is Problem $2B.1$ of Isaacs' Finite Group Theory. You just need to combine elements from the proofs of Theorem $2.8$ and $2.12$ of the same book into one statement.
Induct on $|G|$, $|G|=1$ being trivial. Let $K$ be a proper subgroup of $G$ containing $H$. Then for each $k \in K$ either $\langle H,H^k \rangle$ is nilpotent or $HH^k = H^kH$. So by the induction hypothesis $H \lhd \lhd K$. Suppose $H$ is not subnormal in $G$. Then the Zipper Lemma guarantees the existence of a unique maximal subgroup $M$ with $H \subseteq M$. Let $g \in G$, there are two cases.
$(1)$ $\langle H,H^g \rangle$ is nilpotent. Since every subgroup of a finite nilpotent group is subnormal, $\langle H, H^g \rangle$ is proper in $G$. Therefore $\langle H, H^g \rangle \subseteq M$.
$(2)$ $HH^g=H^gH$. Then $HH^g$ is a subgroup of $G$. But, $H$ is proper and hence $HH^g$ is proper. Thus $HH^g \subseteq M$.
In either case, $H^g \subseteq M$. Thus the normal closure $H^G \subseteq M$, whence $H^G$ is proper in $G$. But then we have $H \lhd \lhd H^G \lhd G$, contradicting $H$ was assumed to be not subnormal in $G$.