Homomorph-containment is not transitive

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 $H\leq K\leq G$ 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 $G$.

Related facts

• Subhomomorph-containment is transitive: Subhomomorph-containment is a stronger subgroup property that is transitive.
• Subhomomorph-containing implies right-transitively homomorph-containing: A homomorph-containing subgroup of a subhomomorph-containing subgroup is homomorph-containing.
• Full invariance is transitive: The property of being a fully invariant subgroup is a weaker subgroup property that is transitive.

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.