Subisomorph-containing not implies homomorph-containing

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 fact about::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
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 $$H$$.