Baer norm is hereditarily permutable

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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

Verbal statement

The Baer norm of any group is a hereditarily permutable subgroup: every subgroup of it is permutable in the whole group.

Statement with symbols

Suppose G is a group and B is the Baer norm of G: B is the intersection of Normalizer (?)s of all the subgroups of G. Then, B is a hereditarily permutable subgroup. In other words, if A is a subgroup of B, and L is a subgroup of G, then A and L permute: AL = LA.

Definitions used

Baer norm

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 G, the Baer norm B(G) is given by:

B(G) = \bigcap_{H \le G} N_G(H).

Hereditarily permutable subgroup

Further information: Hereditarily permutable subgroup

A subgroup H \le G is termed hereditarily permutable if for every subgroup K \le H and every subgroup L \le G, KL = LK, i.e., K and L are permuting subgroups.

Related facts

  • 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).

Proof

Given: A group G with Baer norm B. A subgroup A \le B and a subgroup L \le G.

To prove: AL = LA.

Proof:

  1. A \le N_G(L): Since B is the intersection of normalizers of all subgroups of G, B \le N_G(L). In particular, since A \le B, we have A \le N_G(L).
  2. AL = LA: Since A \le N_G(L), we have aL = La for all a \in A. Thus, AL, which is defined as the union of the aL, a \in A, must equal LA, which is defined as the union of the La, a \in A.