2-subnormality is not finite-join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., 2-subnormal subgroup) not satisfying a subgroup metaproperty (i.e., finite-join-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Join-closed subgroup property (?), .
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about 2-subnormal subgroup|Get more facts about finite-join-closed subgroup propertyGet more facts about join-closed subgroup property|

Statement

A join of two 2-subnormal subgroups of a group need not be 2-subnormal.

Related facts

• 2-subnormal implies join-transitively subnormal: The join of a 2-subnormal subgroup and a $k$-subnormal subgroup is $2k$-subnormal. In particular, the join of two 2-subnormal subgroups is 4-subnormal, and a join of $r$ 2-subnormal subgroups is $2^r$-subnormal.
• Normal implies join-transitively 2-subnormal, and more generally, join of normal and subnormal implies subnormal of same depth: A join of a normal subgroup and a 2-subnormal subgroup is 2-subnormal. More generally, the join of a normal subgroup and a $k$-subnormal subgroup is $k$-subnormal.

Proof

Example of a group of order $64$

We discuss here the example of a group of order 64, given as follows (here $e$ denotes the identity element):

$G=⟨c,b,z,a,u\mid c^4=e,b^2=e,bc{b}^{-1}=c^3,z^2=e,zc{z}^{-1}=c,zb{z}^{-1}=b,a^2=e,ac{a}^{-1}=cz,ab{a}^{-1}=b,u^2=bc,uc{u}^{-1}=c^{3}za,ua{u}^{-1}=c^{2}za⟩$

The two 2-subnormal subgroups are $P=⟨bc,c^2⟩$ and $L=⟨b,c^2⟩$, and their join, which is $H=⟨b,c⟩$, is 3-subnormal but not 2-subnormal.

Note that the GAP ID of $G$ is $(64,32)$: it is the $32^{nd}$ among GAP's list of groups of order $64$.

The lattice of the relevant subgroups of $G$ is depicted in the diagram below:

Let's go over this step by step:

• The subgroup $⟨c,b⟩$ is a dihedral group, which we call $H$. This has three subgroups of order four: the subgroup $K=⟨c⟩$, the subgroup $P=⟨bc,c^2⟩$, and the subgroup $L=⟨b,c^2⟩$. What happens is that $H$ is 3-subnormal, whereas both subgroups $P$ and $L$ are 2-subnormal.
• $H$ is 3-subnormal but not 2-subnormal: In fact, $H$ has a subnormal series $H\le H_1\le H_2\le G$, where $N_G(H)=H_1,N_G(H_1)=H_2,N_G(H_2)=G$. Here, $H_1=⟨b,c,z$ and $H_2=⟨b,c,z,a$. This is both the fastest ascending and the fastest descending subnormal series: the normal closure of $H$ in $G$ is $H_2$, and the normal closure of $H$ in $H_2$ is $H_1$. Note that the presentation is given in a way that fits this subnormal series well.
• $P$ is 2-subnormal: This needs careful understanding: the normalizer of $P$ in $G$ is $H_1=⟨b,c,z⟩$, and this is not normal in $G$. (Thus, $P$ is not a 2-hypernormalized subgroup). Nonetheless, $P$ is 2-subnormal, because the normal closure of $P$, defined as $Q=⟨bc,c^2,z⟩$, is contained in the normalizer.
• $L$ is 2-subnormal: The normalizer of $L$ in $G$ is $H_2$, while its normal closure is the subgroup $N=⟨b,c^2,z,a⟩$. Since the normal closure is contained in the normalizer, $L$ is 2-subnormal. Note that in this case, it is also true that $L$ is a 2-hypernormalized subgroup: the normalizer of $L$ is normal in $G$.