Base of a wreath product implies right-transitively 2-subnormal

Property-theoretic statement
The statement has the following equivalent forms:


 * 1) The subgroup property of being the base of a wreath product is stronger than the subgroup property of being a right-transitively 2-subnormal subgroup.
 * 2) The composition of the subgroup property of being 2-subnormal with the subgroup property of being the base of a wreath product implies the subgroup property of being 2-subnormal.

2-subnormal $$*$$ Base of a wreath product $$\le$$ 2-subnormal

Verbal statement

 * 1) Any base of a wreath product in a group is a right-transitively 2-subnormal subgroup.
 * 2) Any 2-subnormal subgroup of the base of a wreath product is 2-subnormal in the whole group.

Statement with symbols

 * 1) If $$K$$ is the base of a wreath product in $$G$$, then $$K$$ is a right-transitively 2-subnormal subgroup of $$G$$.
 * 2) If $$H$$ is a 2-subnormal subgroup of $$K$$ and $$K$$ is the base of a wreath product in $$G$$, then $$H$$ is 2-subnormal in $$G$$.

Related facts

 * Base of a wreath product is transitive
 * Base of a wreath product implies right-transitively conjugate-permutable

Facts used

 * 1) Direct factor implies transitively normal: Any normal subgroup of a direct factor is normal.

Proof
Given: A group $$G$$ with $$K$$ the base of a wreath product. In other words, $$G = (K \times K \times \dots K) \rtimes L$$ (the direct product may be infinite), and the subgroup $$K$$ is the first direct factor. $$H$$ is a 2-subnormal subgroup of $$K$$. Let $$M = K \times K \times \dots K$$.

To prove: $$H$$ is 2-subnormal in $$G$$.

Proof: Let $$R$$ be the normal closure of $$H$$ in $$K$$.


 * 1) (Fact used: Fact (1), direct factor implies transitively normal): $$R$$ is normal in $$K$$, and $$K$$ is a direct factor of $$M$$. Thus, $$R$$ is normal in $$M$$.
 * 2) (Given data used: $$G$$ is a wreath product of $$K$$ and $$L$$): Since $$R$$ is normal in $$M$$, the normal closure $$S$$ of $$R$$ in $$G$$ is given as the closure of $$R$$ under the action of $$L$$. This is a direct product of isomorphic copies of $$R$$ for all the coordinates in the orbit of the first coordinate.
 * 3) (Given data used: $$H$$ is 2-subnormal in $$K$$): Since $$H$$ is 2-subnormal in $$K$$, $$H$$ is normal in its normal closure $$R$$ in $$K$$.
 * 4) (Fact used: Fact (1), direct factor implies transitively normal): $$H$$ is normal in $$R$$, and $$R$$ is a direct factor of $$S$$, so $$H$$ is normal in $$S$$.
 * 5) Since $$H$$ is normal in $$S$$, and $$S$$ is normal in $$G$$ ($$S$$ is, after all, defined as the normal closure of $$R$$ in $$G$$), $$H$$ is 2-subnormal in $$G$$.