Subhomomorph-containing not implies variety-containing

Statement
It is possible to have a group $$G$$ and a subhomomorph-containing subgroup $$H$$ of $$G$$ that is not a variety-containing subgroup of $$G$$.

Related facts

 * Equivalence of definitions of variety-containing subgroup of finite group: For a finite group, the two notions do coincide, and they also coincide with the notion of subisomorph-containing subgroup.
 * Equivalence of definitions of variety-containing subgroup of periodic group
 * Subisomorph-containing not implies subhomomorph-containing

Proof
Let $$G$$ be the direct product of cyclic groups of order $$n$$ for all positive integers $$n$$. Let $$H$$ be the subgroup of $$G$$ comprising all the elements of finite order.


 * $$H$$ is a subhomomorph-containing subgroup of $$G$$: Any subgroup of $$H$$ is periodic, and the homomorphic image of such a subgroup is thus also periodic. Thus, any homomorphic image of any subgroup of $$H$$ is contained in $$H$$. So, $$H$$ is a subhomomorph-containing subgroup.
 * $$H$$ is not a variety-containing subgroup of $$G$$: $$H$$ contains each of the direct factors of $$G$$, because each factor itself has finite order. Thus, each of these is in the subvariety generated by $$H$$. Hence, so is $$G$$, their direct product. But $$G$$ is not a subgroup of $$H$$, since $$H$$ is a proper subgroup of $$G$$.