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)}
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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 $La,a\in A$.