Subnormality is permuting upper join-closed in finite
From Groupprops
This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup of finite group) satisfying a subgroup metaproperty (i.e., permuting upper 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 of finite group |Get facts that use property satisfaction of subnormal subgroup of finite group | Get facts that use property satisfaction of subnormal subgroup of finite group|Get more facts about permuting upper join-closed subgroup property
Contents
History
The result was proved both by Maier and by Wielandt.
Statement
When the whole group is finite
Suppose is a finite group and
is a subgroup of
. Suppose
are intermediate subgroups of
such that
(i.e., they are Permuting subgroups (?)) and
is a subnormal subgroup in both
and
. Then,
is also subnormal in the product of subgroups
.
Equivalent formulation when the whole group is not finite
Suppose is a group and
is a finite subgroup of
. Suppose
are intermediate finite subgroups of
such that
(i.e., they are Permuting subgroups (?)) and
is a subnormal subgroup in both
and
. Then,
is also subnormal in the product of subgroups
.
Note that these two formulations are equivalent because even if is not finite, the product
is still finite since both
and
are finite.
Related facts
- Subnormality is not permuting upper join-closed
- Subnormality is not finite-upper join-closed (we can get counterexamples even in finite groups)