# Quotient group need not be isomorphic to any subgroup

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., normal subgroup) neednotsatisfy the second subgroup property (i.e., endomorphic kernel)

View a complete list of subgroup property non-implications | View a complete list of subgroup property implications

Get more facts about normal subgroup|Get more facts about endomorphic kernel

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not endomorphic kernel|View examples of subgroups satisfying property normal subgroup and endomorphic kernel

## Contents

## Statement

It is possible to have a group (in fact, we can choose to be a finite group), and a normal subgroup of such that there is no subgroup of isomorphic to the quotient group . In particular, need not be an endomorphism kernel in .

## Related facts

### Opposite facts

- Subgroup lattice and quotient lattice of finite abelian group are isomorphic, and further, under this isomorphism, the corresponding quotient to any subgroup is isomorphic to it. Thus, for a finite abelian group, any quotient group is isomorphic to some subgroup.

### Similar facts

## Proof

### Example of the quaternion group

`Further information: quaternion group, subgroup structure of quaternion group, center of quaternion group`

Suppose is the quaternion group of order eight, and is the center of quaternion group . Then is a Klein four-group. However, all the subgroups of order four in are isomorphic to cyclic group:Z4 and hence, in particular, has no subgroup isomorphic to .