Subnormal not implies conjugate-permutable

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., subnormal subgroup) need not satisfy the second subgroup property (i.e., conjugate-permutable subgroup)
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Statement

Property-theoretic statement

The subgroup property of being a subnormal subgroup does not imply the subgroup property of being conjugate-permutable.

Verbal statement

It is possible to have a subnormal subgroup in a group that is not conjugate-permutable. In fact, it is possible to have a $k$-subnormal subgroup that is not conjugate-permutable for any $k\ge 3$.

Related facts

• 2-subnormal implies conjugate-permutable

Proof

Example of the dihedral group

Consider a dihedral group of order $2^n,n\ge 4$. This is the semidirect product of a cyclic group of order $2^{n-1}$ and a cyclic group of order two, where the latter acts on the former by the inverse map. Here, the cyclic subgroup of order two is a $(n-1)$-subnormal subgroup. We claim that this subgroup is not conjugate-permutable for $n\ge 4$.

To see this, let us make the description of the dihedral group explicit:

$D_{2^n}:=⟨a,x\mid a^{2^{n-1}}=x^2=e,xa{x}^{-1}=a^{-1}⟩$

We claim that the two-element subgroup $\{x,e\}$ is not conjugate-permutable. Indeed, observe that:

$ax{a}^{-1}=x{a}^{-2}$.

So the subgroup $\{x{a}^{-2},e\}$ is a conjugate to the original subgroup. On the other hand, for these two subgroups to permute, we require that $x$ commutes with $a^2$, which is not true since:

$x{a}^2{x}^{-1}=a^{-2}\ne a^2$

for $n\ge 4$.