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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., descendant subgroup) need not satisfy the second subgroup property (i.e., subnormal subgroup)
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Contents 1 Statement 2 Related facts 3 Definitions used 3.1 Descendant subgroup 3.2 Subnormal subgroup 4 Proof 4.1 Example of the group of 2-adic integers

Statement

A descendant subgroup of a group need not be subnormal.

Related facts

• Normality is not transitive
• There exist subgroups of arbitrarily large subnormal depth
• There exist subgroups of arbitrarily large descendant depth
• Ascendant not implies subnormal

Definitions used

Descendant subgroup

Further information: Descendant subgroup

Subnormal subgroup

Further information: Subnormal subgroup

Proof

Example of the group of 2-adic integers

Let K be the group of 2-adic integers under addition. This is the inverse limit of the chain:

.

Let G be the semidirect product of K with a group H of order two, acting via the inverse map.

• H is a descendant subgroup of G: Consider a descending chain K_n defined as follows. K₀ = K, and K_n is the kernel of the quotient map to . Define G_n as the semidirect product of K_n with H. Then, the G_n form a descending chain of subgroups, each having index two in its predecessor, so each is normal in its predecessor. The intersection of all the G_ns is equal to H, and thus, H is a descendant subgroup of G.
• H is not a subnormal subgroup of G: If H were a k-subnormal subgroup of G, then the image of H in G / K_n would be a k-subnormal subgroup of G / K_n for every n. On the other hand, we know that the image of H in G / K_n has subnormal depth exactly n in G / K_n, which is a contradiction for n > k. (For more on this, refer the example of the dihedral group in: there exist subgroups of arbitrarily large subnormal depth).