# Contranormality is upper join-closed

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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 $H \le G$ 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.

## Definitions used

### Contranormal subgroup

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

## Facts used

1. 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.