Baer norm is Dedekind

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 $$B$$.

Related facts

 * Baer norm is hereditarily permutable: Every subgroup of the Baer norm is permutable in the whole group.
 * Baer norm not is hereditarily normal: Every subgroup of the Baer norm need not be normal in the whole group (though this result says that it is normal in the Baer norm).
 * Baer norm is hereditarily 2-subnormal

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 $$B$$.