# Left-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., left-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:

• 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 $H$.