# 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 $H_i \le K_i \le G$ 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.

## Definitions used

### Contranormal subgroup

Further information: 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. 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 $K = L$.