# Descendant not implies subnormal

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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## Statement

A descendant subgroup of a group need not be 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: $\dots \to \mathbb{Z}/2^k\mathbb{Z} \to \mathbb{Z}/2^{k-1}\mathbb{Z} \to \dots \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/\mathbb{Z}$.

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.