2-subnormal implies join-transitively subnormal

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., 2-subnormal subgroup) must also satisfy the second subgroup property (i.e., join-transitively subnormal subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about 2-subnormal subgroup|Get more facts about join-transitively subnormal subgroup

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., 2-subnormal subgroup) must also satisfy the second subgroup property (i.e., linear-bound join-transitively subnormal subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about 2-subnormal subgroup|Get more facts about linear-bound join-transitively subnormal subgroup

Statement

Statement with symbols

Suppose $H,K\leq G$ are subgroups such that $H$ is a 2-subnormal subgroup of $G$ and $K$ is a subnormal subgroup of $G$. Then, the join $\langle H,K\rangle$ is a subnormal subgroup of $G$, and its subnormal depth in $G$ is at most twice the subnormal depth of $K$.

Related facts

• Join of normal and subnormal implies subnormal of same depth
• Permutable and subnormal implies join-transitively subnormal
• Subnormality is normalizing join-closed
• Subnormality is permuting join-closed
• 2-subnormality is conjugate-join-closed

Facts used

1. 2-subnormality is conjugate-join-closed: A join of any collection of 2-subnormal subgroups that are conjugate to each other is again 2-subnormal.
2. Join of subnormal subgroups is subnormal iff their commutator is subnormal: Suppose $H,K$ are subnormal subgroups of a group $G$. Then, consider the subgroups $[H,K]$ (the commutator) of $H$ and $K$), the subgroup $H^{K}$ (the join of $H^{k}$ for all $k\in K$) and the subgroup $\langle H,K\rangle$. If any one of these is subnormal, so are the other two. Further, if $s_{1},s_{2},s_{3}$ denote respectively the subnormal depths of $[H,K],H^{K},\langle H,K\rangle$, we have $s_{3}\leq s_{2}k$.

Proof

Given: A group $G$, a $2$-subnormal subgroup $H$, a $k$-subnormal subgroup $K$.

To prove: $\langle H,K\rangle$ is a $2k$-subnormal subgroup of $G$.

Proof:

Step no.	Assertion	Given data used	Facts used	Previous steps used	Explanation
1	$H^{K}$, i.e., the join (subgroup generated) of conjugates of $H$ by elements of $K$, is a 2-subnormal subgroup of $G$	$H$ is 2-subnormal in $G$	(1)	--	(directly from fact (1))
2	$\langle H,K\rangle$ is $2k$-subnormal	$K$ is $k$-subnormal	(2)	(1)	By step (1), $H^{K}$ is 2-subnormal in $G$. Invoking fact (2), we get that $\langle H,K\rangle$ is also subnormal, and its subnormal depth is bounded by the product of the subnormal depth of $H^{K}$ (which is at most $2$) and the subnormal depth of $K$ (which is $k$). This yields that $\langle H,K\rangle$ is $2k$-subnormal.