Right-transitively homomorph-containing not implies subhomomorph-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., right-transitively homomorph-containing subgroup) need not satisfy the second subgroup property (i.e., subhomomorph-containing subgroup)
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Proof

Let $G$ be the direct product:

$G:=A_{5}\times C_{2}$

where $A_{5}$ is the alternating group of degree five and $C_{2}$ is the cyclic group of order two. Let $H$ be the first direct factor.

Then:

• Every homomorph-containing subgroup of $H$ is a homomorph-containing subgroup of $G$: First, note that $H$ itself is homomorph-containing in $G$, because, since $H$ is simple, no homomorphic image of $H$ can project nontrivially onto the second direct factor. Since $H$ is simple, the only homomorph-containing subgroups of $H$ are $H$ and the trivial subgroup. Both of these are homomorph-containing in <amth>G</math>.
• $H$ is not subhomomorph-containing in $G$: $H$ has cyclic subgroups of order two, that are isomorphic to cyclic subgroups of order two outside $H$, so $H$ is not subhomomorph-containing in $G$.