Left-transitively homomorph-containing not implies subhomomorph-containing

Related facts

 * Fully invariant direct factor implies left-transitively homomorph-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:


 * For any group $$K$$ containing $$G$$ as a homomorph-containing subgroup, $$K$$ also contains $$H$$ as a homomorph-containing subgroup: For any homomorphism $$\alpha:H \to K$$, we can extend it to a homomorphism $$\beta:G \to K$$, since $$H$$ is a direct factor of $$G$$. Since $$G$$ is homomorph-containing in $$K$$, $$\beta(G) \le G$$, so $$\alpha(H) \le G$$. Now, since $$H$$ is homomorph-containing in $$G$$, $$\alpha(H)$$ is contained in $$H$$.
 * $$H$$ is not a subhomomorph-containing subgroup of $$G$$: $$H$$ has cyclic subgroups of order two, that are isomorphic to cyclic subgroups of order two in $$G$$ and outside $$H$$.