Wielandt subgroup is T-group

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., Wielandt subgroup) always satisfies a particular group property (i.e., T-group (?))
View all such group property satisfactions

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.

Definitions used

Wielandt subgroup

Further information: 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

Further information: 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).