Right-transitively homomorph-containing not implies subhomomorph-containing

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 G.
 * $$H$$ 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 $$G$$.