# Homomorph-containment is not transitive

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This article gives the statement, and possibly proof, of a subgroup property (i.e., homomorph-containing subgroup) not satisfying a subgroup metaproperty (i.e., transitive subgroup property).
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## Statement

### Statement with symbols

It is possible to have groups $H \le K \le G$ such that $H$ is a homomorph-containing subgroup of $K$ and $K$ is a homomorph-containing subgroup of $G$, but $H$ is not a homomorph-containing subgroup of $G$.

## Proof

### An example

Consider the group:

$G = SL(2,5) \times \mathbb{Z}/2\mathbb{Z}$.

Let $K$ be the first direct factor $SL(2,5)$ and $H$ be the center of $K$, which is a cyclic subgroup of order two in $K$. Then:

• $K$ is homomorph-containing in $G$: For any homomorphism from $K$ to $G$ the projection to the second direct factor is trivial since $K$ is a perfect group and therefore has no quotient of order two. Thus, any homomorphism from $K$ to $G$ has image in $K$.
• $H$ is homomorph-containing in $K$: In fact, the only element of order two in $K$ is the non-identity element of $H$.
• $H$ is not homomorph-containing in $G$: There is a a homomorphism from $H$ to $G$ mapping $H$ isomorphically to the second direct factor.