Normal not implies left-transitively fixed-depth subnormal

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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., normal subgroup) need not satisfy the second subgroup property (i.e., left-transitively fixed-depth subnormal subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about left-transitively fixed-depth subnormal subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not left-transitively fixed-depth subnormal subgroup|View examples of subgroups satisfying property normal subgroup and left-transitively fixed-depth subnormal subgroup
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., left-transitively fixed-depth subnormal subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about subnormal subgroup|Get more facts about left-transitively fixed-depth subnormal subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property subnormal subgroup but not left-transitively fixed-depth subnormal subgroup|View examples of subgroups satisfying property subnormal subgroup and left-transitively fixed-depth subnormal subgroup

Statement

It is possible to have a group K and a normal subgroup H such that for every k \ge 1, there exists a group G containing K as a k-subnormal subgroup, but in which H is not a k-subnormal subgroup.

Note that this also gives an example where K is a subnormal subgroup of H, and for every k \ge 1, there exists a group G containing K as a k-subnormal subgroup, but in which H is not a k-subnormal subgroup.

Related facts

Proof

Example of the dihedral group

Further information: dihedral group:D8

Let K be the dihedral group of order eight, given by:

K := \langle a,x \mid a^4 = x^2 = 1, xax = a^{-1} \rangle.

Let H be the subgroup of K generated by a^2 and x. H has index two in K and is hence normal. (subgroup of index two is normal).

For any k \ge 1 define G as the group:

G := \langle b,x \mid b^{2^{k+2}} = x^2 = 1, xbx^{-1} = b^{-1} \rangle.

In other words, G is a dihedral group of order 2^{k+3}. Consider K as a subgroup of G by identifying a = b^{2^k} and x = x. Then:

  • The subnormal depth of K in G is k.
  • The subnormal depth of H in G is k + 1.

This completes the proof.