# 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 $La, a \in A$.