Contranormality is UL-join-closed

Statement with symbols
Suppose $$G$$ is a group and $$H_i \le K_i \le G$$ is a family of subgroups indexed by $$i \in I$$, and indexing set. Further, suppose each $$H_i$$ is a contranormal subgroup of $$K_i$$. Then, the join of all the $$H_i$$s is contranormal in the join of all the $$K_i$$s.

Weaker facts

 * Contranormality is upper join-closed

Applications

 * Paranormality is strongly join-closed
 * Polynormality is strongly join-closed
 * Paracharacteristicity is strongly join-closed
 * Polycharacteristicity is strongly join-closed

Contranormal subgroup
Given a subgroup $$H \le K$$ we say that $$H$$ is contranormal in $$K$$ if any normal subgroup of $$K$$ containing $$H$$ must equal the whole of $$K$$. In other words, the normal closure of $$H$$ in $$K$$ is $$K$$.

Facts used

 * 1) uses::Normality satisfies transfer condition: If $$L$$ is normal in $$G$$ and $$K \le G$$, then $$L \cap K$$ is normal in $$K$$.

Proof
Given: Group $$G$$, indexing set $$I$$, $$H_i \le K_i \le G$$ for all $$i \in I$$, and each $$H_i$$ is contranormal in $$K_i$$.

To prove: The join of the $$H_i$$ (which we denote as $$H$$) is contranormal in the join of the $$K_i$$s (which we denote as $$K$$).

Proof: By the definition of contranormality, we need to show that if $$L$$ is a normal subgroup of the join of $$K$$, and $$L$$ contains each $$H$$, then $$L = K$$.

So suppose $$L$$ is normal in $$K$$ with $$H \le L$$. For each $$i \in I$$, $$L \cap K_i$$ is normal in $$K_i$$ by fact (1). Also, $$H_i \le H \le L$$, and $$H_i \le K_i$$ so $$H_i \le L \cap K_i$$. Thus, $$L \cap K_i$$ is a normal subgroup of $$K_i$$ containing $$H_i$$. By contranormality of $$H_i$$ in $$K_i$$, we get $$L \cap K_i = K_i$$, so $$K_i \le L$$. Since this holds for each $$i \in I$$, we get $$K \le L$$, forcing $$K = L$$.