# Subisomorph-containing not implies homomorph-containing

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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) neednotsatisfy the second subgroup property (i.e., homomorph-containing subgroup)

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## Statement

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

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

## 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 be the infinite dihedral group and be the cyclic part:

.

Then:

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