Contranormality is UL-join-closed

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

Statement with symbols

Suppose $G$ is a group and $H_{i}\leq K_{i}\leq G$ 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.

Related facts

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

Definitions used

Contranormal subgroup

Further information: Contranormal subgroup

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

Proof

Given: Group $G$, indexing set $I$, $H_{i}\leq K_{i}\leq G$ 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 $H\leq L$. For each $i\in I$, $L\cap K_{i}$ is normal in $K_{i}$ by fact (1). Also, $H_{i}\leq H\leq L$, and $H_{i}\leq K_{i}$ so $H_{i}\leq L\cap K_{i}$. 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 $K_{i}\leq L$. Since this holds for each $i\in I$, we get $K\leq L$, forcing $K=L$.