Subnormality is normalizing join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) satisfying a subgroup metaproperty (i.e., normalizing join-closed subgroup property)
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Statement

Suppose $H,K\le G$ are subnormal subgroups, with the property that $K\le N_G(H)$: in other words, $K$ normalizes $H$. Then the join of subgroups $⟨H,K⟩$ is also subnormal. Moreover, the subnormal depth of $⟨H,K⟩$ is bounded from above by the products of subnormal depths of $H$ and $K$.

Related facts

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

Facts used

1. Join of normal and subnormal implies subnormal of same depth: If $L$ is normal in $G$ and $K$ is $k$-subnormal in $G$, then $KL$ is subnormal in $G$ with subnormal depth at most $k$.
2. Normality is upper join-closed: If a subgroup is normal in two intermediate subgroups, it is normal in their join.
3. Subnormality satisfies intermediate subgroup condition: More specifically, if $A\le B\le G$ are groups such that $A$ is $k$-subnormal in $G$, then $A$ is also $k$-subnormal in $B$.
4. Subnormal subgroup has a unique fastest descending subnormal series, where the series members are obtained by taking successive normal closures.

Proof

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Given: A group $G$, subnormal subgroups $H,K\le G$ such that $K\le N_G(H)$, i.e., $K$ normalizes $H$. $H$ has subnormal depth $h$ and $K$ has subnormal depth $k$.

To prove: $HK=⟨H,K⟩$ is a subnormal subgroup, with subnormal depth at most $hk$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Consider the descending chain $G_i$ defined by $G_0=G$, and $G_{i+1}$ is the normal closure of $H$ in $G_i$. This is the fastest descending subnormal series for $H$, and thus, $G_h=H$.	Fact (4)	$H$ is $h$-subnormal in $G$
2	$K$ normalizes $G_i$ for all $i$. In particular, for any $i$, $⟨G_i,K⟩=G_{i}K$.		$K$ normalizes $H$	Step (1)	Any subgroup of $G$ defined deterministically in terms of $H$ must be invariant under any automorphism that leaves $H$ invariant.
3	For each $i$, $G_{i+1}$ is normal in $G_{i}K$.	Fact (2)		Steps (1), (2)	By construction, $G_{i+1}$ is normal in $G_i$, and as observed in Step (2), $K$ normalizes $G_{i+1}$, so $G_{i+1}$ is normal in $G_{i}K$ (fact (2)).
4	For each $i$, $K$ is $k$-subnormal in $G_{i}K$	Fact (3)	$K$ is $k$-subnormal in $G$		Given-fact combination direct
5	For each $i$, $G_{i+1}K$ is $k$-subnormal in $G_{i}K$	Fact (1)		Steps (3), (4)	Step-fact combination direct
6	$HK$ is $hk$-subnormal in $G$			Steps (1), (5)	We have a chain: $HK=G_{h}K\le G_{h-1}K\le \dots \le G_{1}K\le G_{0}K=G$ where each member is $k$-subnormal in its successor. This tells us that $HK$ is $hk$-subnormal in $G$.

References

Textbook references

• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, ^{More info}, Page 387, Section 13.1 (Joins and intersections of subnormal subgroups)
• Subnormal subgroups of groups by John C. Lennox and Stewart E. Stonehewer, Oxford Mathematical Monographs, ISBN 019853552X, Page 3, Section 1.2 (First results on joins), Theorem 1.2.1, ^{More info}