# Homomorph-containment is strongly join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., homomorph-containing subgroup) satisfying a subgroup metaproperty (i.e., strongly join-closed subgroup property)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about homomorph-containing subgroup |Get facts that use property satisfaction of homomorph-containing subgroup | Get facts that use property satisfaction of homomorph-containing subgroup|Get more facts about strongly join-closed subgroup property

## Statement

### Statement with symbols

Suppose $G$ is a group, $I$ is an indexing set, and $H_i, i \in I$, is a collection of homomorph-containing subgroups of $G$. Then, the join of subgroups $\langle H_i \rangle_{i \in I}$ is also a homomorph-containing subgroup of $G$.

## Related facts

Given: A group $G$, an indexing set $I$, a collection $H_i$ of homomorph-containing subgroups of $G$, $i \in I$. $H = \langle H_i \rangle_{i \in I}$. A homomorphism $\varphi: H \to G$.
To prove: $\varphi(H)$ is containined in $H$.
Proof: $\varphi(H) = \varphi \langle H_i \rangle_{i \in I} = \langle \varphi(H_i) \rangle_{i \in I}$. Since each $H_i$ is homomorph-containing in $G$, $\varphi(H_i)$ is contained in $H_i$, so the join of $\varphi(H_i), i \in I$, is contained in the join of the $H_i$s, which is $H$.