Subnormality is normalizing join-closed: Difference between revisions

Latest revision as of 14:39, 3 July 2014

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)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about subnormal subgroup |Get facts that use property satisfaction of subnormal subgroup | Get facts that use property satisfaction of subnormal subgroup|Get more facts about normalizing join-closed subgroup property

Statement

Suppose $H, K \leq G$ are subnormal subgroups, with the property that $K \leq N_{G} (H)$ : 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

Facts used

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

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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

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

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	$H K$ is $h k$ -subnormal in $G$			Steps (1), (5)	We have a chain: $H K = G_{h} K \leq G_{h - 1} K \leq \dots \leq G_{1} K \leq G_{0} K = G$ where each member is $k$ -subnormal in its successor. This tells us that $H K$ is $h k$ -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}

@@ Line 18: / Line 18: @@
 # [[uses::Normality is upper join-closed]]: If a subgroup is normal in two intermediate subgroups, it is normal in their join.
 # [[uses::Subnormality satisfies intermediate subgroup condition]]: More specifically, if <math>A \le B \le G</math> are groups such that <math>A</math> is <math>k</math>-subnormal in <math>G</math>, then <math>A</math> is also <math>k</math>-subnormal in <math>B</math>.
+# [[uses::Subnormal subgroup has a unique fastest descending subnormal series]], where the series members are obtained by taking successive [[normal closure]]s.
 ==Proof==
@@ Line 32: / Line 33: @@
 ! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
 |-
-| 1 || Consider the descending chain <math>G_i</math> defined by <math>G_0 = G</math>, and <math>G_{i+1}</math> is the normal closure of <math>H</math> in <math>G_i</math>. This is the fastest descending subnormal series for <math>H</math>, and thus, <math>G_h = H</math>. || (general theory of subnormal subgroups) || <math>H</math> is <math>h</math>-subnormal in <math>G</math> || ||
+| 1 || Consider the descending chain <math>G_i</math> defined by <math>G_0 = G</math>, and <math>G_{i+1}</math> is the normal closure of <math>H</math> in <math>G_i</math>. This is the fastest descending subnormal series for <math>H</math>, and thus, <math>G_h = H</math>. || Fact (4) || <math>H</math> is <math>h</math>-subnormal in <math>G</math> || ||
 |-
 | 2 || <math>K</math> normalizes <math>G_i</math> for all <math>i</math>. In particular, for any <math>i</math>, <math>\langle G_i, K \rangle = G_iK</math>. || || <math>K</math> normalizes <math>H</math> || Step (1) || Any subgroup of <math>G</math> defined deterministically in terms of <math>H</math> must be invariant under any automorphism that leaves <math>H</math> invariant.