2-subnormality is not finite-join-closed

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.

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 $$G$$.