2-subnormal not implies hypernormalized

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., 2-subnormal subgroup) need not satisfy the second subgroup property (i.e., hypernormalized subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about 2-subnormal subgroup|Get more facts about hypernormalized subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property 2-subnormal subgroup but not hypernormalized subgroup|View examples of subgroups satisfying property 2-subnormal subgroup and hypernormalized subgroup

Statement

Verbal statement

A 2-subnormal subgroup of a group need not be hypernormalized.

Related facts

Stronger facts

• Abnormal normalizer and 2-subnormal not implies normal: In fact, our first example used here works for that as well, and our second construction can be specialized to an example for that as well.

Similar facts

• 3-subnormal subgroup need not have a unique fastest ascending subnormal series uses a generic construction that is similar to the second construction in this page.

Proof

An example in the symmetric group on four letters

Let $G$ be the symmetric group on four letters $\{1,2,3,4\}$ and $H$ be the two-element subgroup generated by $(13)(24)$.

Then, $H$ is normal in the subgroup $K=\{(),(12)(34),(13)(24),(14)(23)\}$ (V4 in S4), which is normal in $G$. So $H$ is 2-subnormal in $G$.

On the other hand, the normalizer $N_G(H)$ is a dihedral subgroup of order eight, which is a self-normalizing subgroup. So, $H$ is not hypernormalized in $G$.

A general construction

Let $G$ be a group and $H$ be a proper self-normalizing subgroup of $G$ (such examples exist; for instance D8 in S4 mentioned in the preceding example).

Let $p$ be a prime. Let $A$ be the wreath product of the group of prime order with $G$. Equivalently, $A={\mathbb{F}}_p[G]⋊G$ is the semidirect product of the additive group of ${\mathbb{F}}_p[G]$ with $G$ acting by left multiplication.

Let $B={\mathbb{F}}_p[H]$.

Then, $B$ is 2-subnormal in $A$, because the following 2-subnormal series works:

${\mathbb{F}}_p[H]\le {\mathbb{F}}_p[G]\le {\mathbb{F}}_p[G]⋊G=A$

The key point here is that the normality at the first step has nothing to do with what kind of subgroup (or subset!) $H$ is in $G$; it can be seen just from ${\mathbb{F}}_p[G]$ being an abelian group and ${\mathbb{F}}_p[H]$ being a subgroup.

On the other hand, $N_A(B)={\mathbb{F}}_p[G]⋊H$, and from the assumption that $H$ is self-normalizing in $G$, we get that $N_A(B)$ is self-normalizing in $A$.

Therefore, $B$ is not hypernormalized in $A$, and we therefore have an example ($B$ in $A$) of a 2-subnormal subgroup that is not hypernormalized.