Wielandt subgroup is T-group

Statement
The Wielandt subgroup of a group (defined as the intersection of the normalizers of its subnormal subgroups) is a T-group: every subnormal subgroup of it is normal.

Note that this is the strongest group property true for the Wielandt subgroup, because every T-group equals its own Wielandt subgroup.

Wielandt subgroup
For a group $$G$$, the Wielandt subgroup $$W(G)$$ is defined as the intersection of the normalizers of all the subnormal subgroups of $$G$$.

T-group
A group is termed a T-group if every subnormal subgroup of the group is normal.

Proof
Given: A group $$G$$ with Wielandt subgroup $$W$$. A subnormal subgroup $$A$$ of $$W$$.

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

Proof:


 * 1) $$W$$ is characteristic in $$G$$: Any automorphism of $$G$$ sends subnormal subgroups to subnormal subgroups; hence, it sends normalizers of subnormal subgroups to normalizers of subnormal subgroups. In particular, it sends the intersection of these to the intersection of these, so any automorphism of $$G$$ preserves $$W$$.
 * 2) $$A$$ is subnormal in $$G$$: Since $$A$$ is subnormal in $$W$$, and $$W$$ is characteristic, hence normal, in $$G$$, we obtain that $$A$$ is subnormal in $$G$$.
 * 3) $$W$$ is contained in the normalizer of $$A$$: This follows from the fact that $$A$$ is subnormal in $$G$$, and that the Wielandt subgroup is contained in the normalizers of all subnormal subgroups.
 * 4) $$A$$ is normal in $$W$$: This follows directly from step (3).