Contranormality is upper join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., contranormal subgroup) satisfying a subgroup metaproperty (i.e., upper join-closed subgroup property)
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Statement

Statement with symbols

Suppose $H\leq G$ is a subgroup, and $K_{i},i\in I$ , is an indexed family of subgroups with $H\leq K_{i}$ 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.

Definitions used

Contranormal subgroup

Further information: contranormal subgroup

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

Related facts

Stronger facts

Contranormality is UL-join-closed

Applications

Paranormal implies polynormal

Facts used

Normality satisfies transfer condition: If $L\triangleleft G$ is a normal subgroup, and $K\leq G$ , then $L\cap K$ is normal in $K$ .

Proof

Given: $H\leq G$ , family of subgroups $K_{i},i\in I$ with $H\leq K_{i}$ , 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 $K_{i}\leq L$ for each $i\in I$ . Thus, $L$ must equal the join of the $K_{i}$ s.