Descendant subgroup

This article defines a subgroup property: a property that can be evaluated to true/false given a group and a subgroup thereof, invariant under subgroup equivalence. View a complete list of subgroup properties[SHOW MORE]
If the ambient group is a finite group, this property is equivalent to the property: subnormal subgroup
View other properties finitarily equivalent to subnormal subgroup | View other variations of subnormal subgroup |
This is a variation of subnormality|Find other variations of subnormality |

Definition

A subgroup $H$ of a group $G$ is termed descendant if we have subgroups $H_\alpha$ of $G$ for every ordinal $\alpha$ such that:

• $H_0 = G$
• $H_{\alpha + 1} \ \underline{\triangleleft} \ H_\alpha$ (i.e., $H_{\alpha + 1}$ is a normal subgroup of $H_\alpha$) for every ordinal $\alpha$.
• If $\alpha$ is a limit ordinal, then $H_\alpha = \bigcap_{\gamma < \alpha} H_\gamma$.

and such that there is some ordinal $\beta$ such that $H_\beta = H$.

In terms of the descendant closure operator

The subgroup property of being an descendant subgroup is obtained by applying the descendant closure operator to the subgroup property of being normal.

Facts

Descendant-contranormal factorization

This result states that given any subgroup $H$ of $G$, there is a unique subgroup $K$ containing $H$ such that $H$ is contranormal in $K$ and $K$ is descendant in $G$.

Metaproperties

Metaproperty name Satisfied? Proof Statement with symbols
transitive subgroup property Yes descendance is transitive If $H \le K \le G$ are groups such that $H$ is a descendant subgroup of $K$ and $K$ is a descendant subgroup of $G$, then $H$ is a descendant subgroup of $G$.
trim subgroup property Yes Every group is descendant in itself, and the trivial subgroup is descendant in any group.
intermediate subgroup condition Yes descendance satisfies intermediate subgroup condition If $H \le K \le G$ are groups such that $H$ is descendant in $G$, then $H$ is descendant in $K$.
strongly intersection-closed subgroup property Yes descendance is strongly intersection-closed If $H_i, i \in I$, are all descendant subgroups of $G$, so is the intersection $\bigcap_{i \in I} H_i$.
image condition No descendance does not satisfy image condition It is possible to have groups $G$ and $K$, a descendant subgroup $H$ of $G$ and a surjective homomorphism $\varphi:G \to K$ such that $\varphi(H)$ is not a descendant subgroup of $K$.