Join of subnormal subgroups is subnormal iff their commutator is subnormal
Statement
Suppose are Subnormal subgroup (?)s of a group . Then, the following are equivalent:
- The join of subgroups is a subnormal subgroup of .
- The subgroup , i.e., the closure of under the action of by conjugation, is a subnormal subgroup of .
- The commutator of two subgroups is a subnormal subgroup of .
Moreover, if denote the subnormal depths of the subgroups in (1),(2),(3) respectively, and denote the subnormal depths of , we have:
.
Related facts
Facts used
- Commutator of two subgroups is normal in join: For , is normal in . Further, we have , , and .
- Subnormality is normalizing join-closed: For subnormal of depths , with , we have that is subnormal of depth at most .
Proof
Given: A group with subnormal subgroups .
To prove: Consider the subgroups , , and . If any one of them is subnormal, so are the others. Further, if are their subnormal depths, we have:
.
Bounding the commutator in terms of the join
To prove: If is subnormal of depth , so is , and if denotes its subnormal depth, we have .
Proof: By fact (1), the commutator is a normal subgroup of , and thus, if is subnormal of depth , we have a subnormal series for of length . This yields .
Bounding the commutator in terms of the closure
To prove: If is subnormal of depth , then is subnormal of depth with .
Proof: By fact (1), is normal in , and in particular, is normal in the intermediate subgroup . Thus, if has subnormal depth , we obtain a subnormal series for of length , yielding .
Bounding the closure in terms of the join
To prove: If is subnormal of depth , then is subnormal of depth with .
Proof: We have , and by definition, normalizes . Thus, is normal in . If is subnormal of depth , we get a subnormal series for of length , yielding is subnormal with subnormal depth at most .
Bounding the closure in terms of the commutator
To prove: If is subnormal of depth and is subnormal of depth , then is subnormal of depth at most .
Proof: By fact (1), normalizes and . Thus, by fact (2), is subnormal of depth at most .
Bounding the join in terms of the closure
To prove: If is subnormal of depth and is subnormal of depth , then is subnormal of depth at most .
Proof: We have , with normalizing . Thus, by fact (2), is subnormal of depth at most .
References
Textbook references
- A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, ^{More info}, Page 388, Theorem 13.1.6, Section 13.1 (Joins and intersections of subnormal subgroups)