Subnormality is permuting join-closed

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

Verbal statement

A join of two subnormal subgroups that permute is again a subnormal subgroup. Moreover, its subnormal depth is bounded by a function of the subnormal depths of the individual subnormal subgroups.

Statement with symbols

Suppose $H$ and $K$ are subnormal subgroups of a group $G$, and suppose further that $HK=KH$, i.e., the product of subgroups is again a subgroup. Then, $HK$ is a subnormal subgroup of $G$. Further, if the subnormal depth of $H$ is $h$ and the subnormal depth of $K$ is $k$, the subnormal depth of $HK$ is bounded by:

$h[k]^{h}=hk(k+1)(k+2)\dots (k+h-1)$.

By symmetry, it is also bounded by the function $k[h]^{k}$. When $h\leq k$, $h[k]^{h}$ is the smaller of the two numbers, hence the more relevant bound.

Related facts

• Subnormality is normalizing join-closed
• Normality is strongly join-closed
• 2-subnormality is conjugate-join-closed
• Join of normal and subnormal implies subnormal of same depth
• Commutator subgroup satisfies ascending chain condition on subnormal subgroups implies subnormal join property
• Nilpotent commutator subgroup implies subnormal join property

Applications

• Permutable and subnormal implies join-transitively subnormal

Examples

Subnormal depth $h$ of $H$	Subnormal depth $k$ of $K$	Upper bound on subnormal depth of $HK$ given by this statement	Best known bound on subnormal depth of $HK$	Lowest possible upper bound based on explicit example
0	anything	0	0	0
1	$k$	$k$	$k$	$k$ -- take $H$ trivial.
2	2	12	4 (see 2-subnormal implies join-transitively subnormal)	3 (see 2-subnormality is not finite-join-closed)
2	$k$	$2k(k+1)$	$2k$ (see 2-subnormal implies join-transitively subnormal)	?
3	3	180	180	?
3	4	360	360	?

Facts used

1. Modular property of groups: If $A,B,C\leq G$ such that $A\leq C$ we have $A(B\cap C)=AB\cap C$.
2. Subnormality is normalizing join-closed: If $A,B\leq G$ are subnormal of depths $a,b$, and $B\leq N_{G}(A)$, then $AB$ is a subnormal subgroup of depth at most $ab$.
3. Subnormality satisfies intermediate subgroup condition: If $H\leq L\leq G$ and $H$ is $h$-subnormal in $G$, $H$ is $h$-subnormal in $L$.
4. Subnormality is finite-relative-intersection-closed: Suppose $A$ is $a$-subnormal in $C$ and $B\leq C$ such that $B$ is $b$-subnormal in some subgroup $D$ of $C$ containing both $A$ and $B$. Then, $A\cap B$ is $(a+b)$-subnormal in $C$.

Proof

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Given: A group $G$, a subnormal subgroup $H$ of subnormal depth $h$ in $G$, a subnormal subgroup $K$ of subnormal depth $k$, such that $HK=KH$. Let $J=HK$.

To prove: $J$ is a subnormal subgroup of $G$ with subnormal depth at most $hk(k+1)\dots (k+h-1)$.

Proof: We break the proof into many steps. Note that we use permutability where we use the fact that the product $HK$ is the subgroup generated by $H$ and $K$.

The key idea is to construct one subnormal series, first going down the series, and then going back up the series using some bounds. Our first few steps are in the ambient group $J$, and our later steps shift focus to the ambient group $G$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation	Commentary
1	$H$ is $h$-subnormal in $J$	Fact (3)	$H$ is $h$-subnormal in $G$	--	Piece together.	Shifting to ambient group $J$.
2	Define a subnormal series for $H$ in $J$ by $H_{0}=J$, and $H_{i+1}=H^{H_{i}}$. Note that $H_{0}\geq H_{1}\geq H_{2}\geq \dots$					Still in ambient group $J$, viewing subnormality there.
3	$H_{h}=H$. Moreover, each $H_{i}$ is $i$-subnormal in $J$.			Steps (1), (2)	This follows from the definition of subnormal depth.	Still in ambient group $J$, viewing subnormality there.
4	$H_{i}\cap K$ is $(k+i)$-subnormal in $G$	Fact (4)	$K$ is $k$-subnormal in $G$	Step (3)	Piece together. In Fact (4), set $A=K,B=H_{i},C=G,D=J$.	Crucial step, where we switch to ambient group $G$.
5	$H_{i}=H_{i+1}(K\cap H_{i})$	Fact (1)	$H$ and $K$ permute	Step (2)	[SHOW MORE] By fact (1), applied to $A=H_{i+1},B=K,C=H_{i}$, we have $H_{i+1}(K\cap H_{i})=H_{i+1}K\cap H_{i}$. But we know that $H\leq H_{i+1}$, so $HK\leq H_{i+1}K$, so $J\leq H_{i+1}K$, so $H_{i}\leq H_{i+1}K$. Thus, we get $H_{i+1}(K\cap H_{i})=H_{i}$.	Preparing the ground for moving up from $H$ to $J$ via $H_{i}$s in ambient group $G$.
6	$K\cap H_{i}$ normalizes $H_{i+1}$			Step (2)	[SHOW MORE] By definition of $H_{i+1}$, we know that $H_{i+1}$ is normal in $H_{i}$, so $H_{i}$ normalizes $H_{i+1}$. Thus, the subgroup $K\cap H_{i}$ normalizes $H_{i+1}$.	Preparing the ground for moving up from $H$ to $J$ via $H_{i}$s in ambient group $G$.
7	If $H_{i+1}$ has subnormal depth $s$, then $H_{i}$ is subnormal of depth at most $s(k+i)$	Fact (2)	--	Step (4)	[SHOW MORE] By step (4), $H_{i}\cap K$ is subnormal in $G$ of depth at most $k+i$. By step (6), $K\cap H_{i}$ normalizes $H_{i+1}$. Thus, by fact (2), the join of these subgroups is subnormal and its subnormal depth is at most $s(k+i)$.	Single step up the ladder from $H$ to $J$ via $H_{i}$s, in ambient group $G$.
8	The final result		$H$ is $h$-subnormal in $G$)	Steps (2), (7)	[SHOW MORE] Step (7), and iterating downward from $i=h-1$ to $i=0$, yields the required result. Note that the first step, i.e. $i=h-1$, uses the fact that $H$ is $h$-subnormal in $G$.	Combine multiplicative factors accumulated in each step of the climb up.

References

Textbook references

• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, ^{More info}
• Subnormal subgroups of groups by John C. Lennox and Stewart E. Stonehewer, Oxford Mathematical Monographs, ISBN 019853552X, Page 5, Section 1.2 (First results on joins), Theorem 1.2.5, ^{More info}