Linear-bound join-transitively subnormal subgroup

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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]

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If the ambient group is a finite group, this property is equivalent to the property: subnormal subgroup
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Definition

A subgroup $H$ of a group $G$ is termed linear-bound join-transitively subnormal if there exists a natural number $n$ such that, given any $k$-subnormal subgroup $K$ of $G$, the join $\langle H,K\rangle$ is $nk$-subnormal.

Here, a $k$-subnormal subgroup is a subgroup whose subnormal depth is at most $k$.

Relation with other properties

Stronger properties

• Normal subgroup: For a normal subgroup, we can set $n=1$. For full proof, refer: Join of normal and subnormal implies subnormal of same depth
• 2-subnormal subgroup: For a 2-subnormal subgroup, we can set $n=2$. For full proof, refer: 2-subnormal implies join-transitively subnormal
• Subnormal subgroup of finite index
• Intermediately linear-bound join-transitively subnormal subgroup
• Linear-bound intermediately join-transitively subnormal subgroup
• Asymptotically fixed-depth join-transitively subnormal subgroup

Weaker properties

• Polynomial-bound join-transitively subnormal subgroup
• Join-transitively subnormal subgroup
• Subnormal subgroup

Metaproperties

Transitivity

NO: This subgroup property is not transitive: a subgroup with this property in a subgroup with this property, need not have the property in the whole group
ABOUT THIS PROPERTY: View variations of this property that are transitive|View variations of this property that are not transitive
ABOUT TRANSITIVITY: View a complete list of subgroup properties that are not transitive|View facts related to transitivity of subgroup properties | View a survey article on disproving transitivity

This follows from the fact that normal subgroups are linear-bound join-transitively subnormal, but not all subnormal subgroups are linear-bound join-transitively subnormal. That in turn follows from the fact that subnormality is not finite-join-closed. For full proof, refer: Join of two 3-subnormal subgroups need not be subnormal, subnormality is not finite-join-closed

If $H_{1},H_{2}\leq G$ are both linear-bound join-transitively subnormal with corresponding natural numbers $n_{1},n_{2}$, their join is linear-bound join-transitively subnormal with natural number $n_{1}n_{2}$. (A smaller natural number might also work for the join).