Subisomorph-containing not implies homomorph-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., subisomorph-containing subgroup) need not satisfy the second subgroup property (i.e., homomorph-containing subgroup)
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Statement

It is possible to have a group $G$ and a subgroup $H$ of $G$ such that $H$ is a subisomorph-containing subgroup of $G$ (i.e., it contains any subgroup of $G$ isomorphic to a subgroup of $H$), but $H$ is not a homomorph-containing subgroup: there is a homomorphic image of $H$ in $G$ that is not contained in $H$.

In particular, this also shows that $H$ is not a Subhomomorph-containing subgroup (?) of $G$.

Related facts

• Equivalence of definitions of variety-containing subgroup of finite group: This states that in a finite group, being a subisomorph-containing subgroup, a subhomomorph-containing subgroup, and a variety-containing subgroup, are all equivalent.

Proof

Further information: infinite dihedral group

Let $G$ be the infinite dihedral group and $H$ be the cyclic part:

$G=\langle a,x\mid x^{2}=e,xax=a^{-1}\rangle ,\qquad H=\langle a\rangle$.

Then:

• $H$ is a subisomorph-containing subgroup of $G$: Every subgroup of $G$ contained in $H$ is either trivial or infinite cyclic. On the other hand, every element in $G$ and outside $H$ has order two. Thus, no subgroup of $G$ outside $H$ is isomorphic to a subgroup of $H$.
• $H$ is not a homomorph-containing subgroup of $G$: $H$ is infinite cyclic and the group $\langle x\rangle$, which is cyclic of order two in $G$ and not contained in $H$, is a homomorphic image of $H$.