Baer norm is hereditarily permutable

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 fact about::normalizers 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: $$AL = LA$$.

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
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 $$La, a \in A$$.