Join of two 3-subnormal subgroups may be proper and contranormal
From Groupprops
Contents
Statement
It is possible to have a group and two 3-subnormal subgroup (?)s and of such that the join is not a Subnormal subgroup (?) of . In fact, it can happen that the join is a proper Contranormal subgroup (?).
Related facts
Similar facts
- Subnormality is not finite-join-closed is an easy corollary.
- 3-subnormality is not finite-join-closed is another easy corollary.
- 3-subnormal not implies finite-automorph-join-closed subnormal: This is not a corollary of the statement, but the same example works.
- 4-subnormal not implies finite-conjugate-join-closed subnormal
Opposite facts
- Join of normal and subnormal implies subnormal of same depth
- 2-subnormal implies join-transitively subnormal
- 3-subnormal implies finite-conjugate-join-closed subnormal
Proof
Construction of the counterexample
The construction involves the following steps:
- Let be the set of all subsets of such that there exists integers such that contains all integers less than and no integer greater than .
- Let be an elementary abelian -group with basis (as a vector space over the field of two elements) given by , where ranges over .
- Let be an elementary abelian -group with basis (as a vector space over the field of two elements) given by , where ranges over . (Note that and are isomorphic).
- Let be the direct product of and .
- For every , define if , and if . Analogously, define . Now define, for :
- Automorphisms given on the basis by .
- Automorphisms given on the basis by .
- Let be the subgroup of generated by the and be the subgroup of generated by the .
- Define , again as a subgroup of .
- Define as the external semidirect product , with the action of the usual action by automorphisms.
Then, both and are -subnormal subgroups of , but is not a subnormal subgroup of .
Preliminary computations
Claim: and .
Proof: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]Proof that and are both -subnormal
We prove that is -subnormal in three steps:
- The normal closure of in is : PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
- The normal closure of in is : PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
- is normal in : In fact, is an internal direct product of and .
Proof that is proper and contranormal
The normal closure of in contains both and . Thus, the normal closure of in contains , and hence must be the whole group .
That is proper follows because it is the non-normal part of a semidirect product with a nontrivial group .
References
Textbook references
- A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, Page 389, ^{More info}