2-subnormality is not finite-join-closed

From Groupprops
Jump to: navigation, search
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

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:

2subnormalityisnotfinitejoinclosed.png

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.