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Definition 1: A subgroup $H$ of a group $G$ is said to be abnormal in $G$ if for each $g\in G$, we have $g\in \langle H, H^g \rangle$.

Definition 2: A finite group $G$ is called a $B$-group if every $p$-subgroup of $G$ (for a prime $p$) is either normal or abnormal in $G$

Groups of order $pq$ where $p, q$ are primes are examples of $B$-groups.

Proposition: A $B$-group is a solvable $T$-group i.e. a solvable group in which normality is transitive. In particular, $B$-groups are supersolvable.

I'm not sure if the converse i.e. a finite supersolvable group is a $B$-group, holds in general, as it is not mentioned by the author of the paper here. I tried to prove the assertion but it eludes me. The next best thing would be finding a counterexample of a finite supersolvable group $G$ with some $p$-subgroup $H$ that is neither normal nor abnormal in $G$. Any help with finding this counterexample would be greatly appreciated.

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    $\begingroup$ How about the dihedral group of order $8$? $\endgroup$
    – Derek Holt
    Commented Sep 5, 2017 at 19:10
  • $\begingroup$ Indeed @DerekHolt, thank you. The non-normal subgroup, say $H$ of $D_8$ order 2, which is isomorphic to $\mathbb{Z_2}$, has normaliser, $N_{D_8} (H) = V_4$, which is the Klein 4-group. Thus $H$ cannot be self-normalisiing, and consequently not abnormal in $G$. $\endgroup$
    – R Maharaj
    Commented Sep 6, 2017 at 6:19

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