# 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 Subhomomorph-containing subgroup (?) of $G$.

## 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 $H$.