Baer norm is hereditarily permutable
This article gives the statement, and possibly proof, of the fact that for any group, the subgroup obtained by applying a given subgroup-defining function (i.e., Baer norm) always satisfies a particular subgroup property (i.e., hereditarily permutable subgroup)}
View subgroup property satisfactions for subgroup-defining functions View subgroup property dissatisfactions for subgroup-defining functions
Statement with symbols
Suppose is a group and is the Baer norm of : is the intersection of Normalizer (?)s of all the subgroups of . Then, is a hereditarily permutable subgroup. In other words, if is a subgroup of , and is a subgroup of , then and permute: .
Further information: Baer norm
The Baer norm of a group is defined as the intersection of normalizers of all its subgroups. In symbols, for a group , the Baer norm is given by:
Hereditarily permutable subgroup
Further information: Hereditarily permutable subgroup
A subgroup is termed hereditarily permutable if for every subgroup and every subgroup , , i.e., and are permuting subgroups.
- Baer norm not is hereditarily normal: The Baer norm of a group need not be hereditarily normal: it may have subgroups that are not normal in the whole group.
- Baer norm is Dedekind: Every subgroup of the Baer norm is normal inside the Baer norm (though, as pointed above, it need not be normal in the whole group).
Given: A group with Baer norm . A subgroup and a subgroup .
To prove: .
- : Since is the intersection of normalizers of all subgroups of , . In particular, since , we have .
- : Since , we have for all . Thus, , which is defined as the union of the , must equal , which is defined as the union of the .