# 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 (?), .
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## Statement

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

$G = \langle c,b,z,a,u \mid c^4 = e, b^2 = e, bcb^{-1} = c^3, z^2 = e, zcz^{-1} = c, zbz^{-1} = b, a^2 = e, aca^{-1} = cz, aba^{-1} = b, u^2 = bc, ucu^{-1} = c^3za, uau^{-1} = c^2za \rangle$

The two 2-subnormal subgroups are $P = \langle bc, c^2 \rangle$ and $L = \langle b, c^2 \rangle$, and their join, which is $H = \langle b, c \rangle$, 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 $\langle c,b \rangle$ is a dihedral group, which we call $H$. This has three subgroups of order four: the subgroup $K = \langle c \rangle$, the subgroup $P = \langle bc, c^2 \rangle$, and the subgroup $L = \langle b, c^2 \rangle$. 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, $H_1 = \langle b,c,z$ and $H_2 = \langle b,c,z,a$. 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 $H_1 = \langle b,c,z \rangle$, 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 $Q = \langle bc, c^2, z \rangle$, 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 $N = \langle b,c^2,z,a \rangle$. 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 $G$.