Descendant not implies subnormal

Statement
A descendant subgroup of a group need not be subnormal.

Related facts

 * Stronger than::Normality is not transitive
 * Stronger than::There exist subgroups of arbitrarily large subnormal depth
 * Weaker than::There exist subgroups of arbitrarily large descendant depth
 * Ascendant not implies subnormal

Related facts

 * Ascendant not implies subnormal
 * Descendance does not satisfy image condition

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:

$$\dots \to \mathbb{Z}/2^k\mathbb{Z} \to \mathbb{Z}/2^{k-1}\mathbb{Z} \to \dots \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/\mathbb{Z}$$.

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. $$K_0 = K$$, and $$K_n$$ is the kernel of the quotient map to $$\mathbb{Z}/2^n\mathbb{Z}$$. 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 $$G_n$$s 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).

Example of the infinite dihedral group
Suppose $$G$$ is the infinite dihedral group with cyclic maximal subgroup (isomorphic to the group of integers) $$C$$ and a complementary subgroup $$H$$ of order two. Then, $$H$$ is a descendant subgroup, as it is the intersection of the descending chain:

$$G = C \rtimes H \ge 2C \rtimes H \ge 4C \rtimes H \dots$$

Note that each member has index two in its predecessor, hence is normal in its predecessor.

This example is very similar in nature to the 2-adic example.