Subhomomorph-containing not implies variety-containing

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subhomomorph-containing subgroup) need not satisfy the second subgroup property (i.e., variety-containing subgroup)
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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 $G$.