Contranormality is upper join-closed

Statement with symbols
Suppose $$H \le G$$ is a subgroup, and $$K_i, i \in I$$, is an indexed family of subgroups with $$H \le K_i$$ for each $$i \in I$$. Then, if $$H$$ is contranormal in each $$K_i$$, $$H$$ is also contranormal in the [join of subgroups|join]] of the $$K_i$$s.

Contranormal subgroup
$$H \le K$$ is a contranormal subgroup if for any $$L \le K$$ containing $$H$$ such that $$L$$ is normal in $$K$$, $$L = K$$.

Stronger facts

 * Contranormality is UL-join-closed

Applications

 * Paranormal implies polynormal

Facts used

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

Proof
Given: $$H \le G$$, family of subgroups $$K_i, i \in I$$ with $$H \le K_i$$, and $$H$$ contranormal in each $$K_i$$.

To prove: $$H$$ is normal in the join of all the $$K_i$$s.

Proof: Suppose $$L$$ is a normal subgroup of the join of the $$K_i$$s, containing $$H$$. Then, for each $$K_i$$, $$L \cap K_i$$ is a subgroup of $$K_i$$ containing $$H$$, and by fact (1), it is normal in $$K_i$$. Since $$H$$ is contranormal in $$K_i$$, $$L \cap K_i = K_i$$ for each $$i \in I$$, so $$K_i \le L$$ for each $$i \in I$$. Thus, $$L$$ must equal the join of the $$K_i$$s.