Baer norm is Dedekind

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 group property (i.e., Dedekind group (?))
View all such group property satisfactions

Statement

Verbal statement

The Baer norm of a group, defined as the intersection of the normalizers of all its subgroups, is a Dedekind group: every subgroup of it is normal in it.

Note that this is the strongest condition we can put on the Baer norm, because every Dedekind group occurs as its own Baer norm.

Statement with symbols

If $G$ is a group and $B$ is the Baer norm of $G$, then for any subgroup $A \le B$, $A$ is normal in $B$.

Proof

Given: A group $G$ with Baer norm $B$. A subgroup $A \le B$.

To prove: $A$ is normal in $B$.

Proof: Since $B$ is the intersection of normalizers of all subgroups of $G$, $B \le N_G(A)$. Thus, $A$ is normal in $B$.