# Subgroup need not be isomorphic to any quotient group

## Contents

## Statement

It is possible to have a group (in fact, we can choose to be a finite group) and a Subgroup (?) of such that there is no normal subgroup of for which is isomorphic to the Quotient group (?) .

## Related facts

### Opposite facts

- Subgroup lattice and quotient lattice of finite abelian group are isomorphic and in particular, this implies that for a finite abelian group, every subgroup is isomorphic to some quotient group, and vice versa.

### Similar facts

## Proof

### Example of symmetric group of degree three

`Further information: symmetric group:S3, subgroup structure of symmetric group:S3, A3 in S3`

Let be the symmetric group of degree three (order six) acting on the set and be the unique subgroup of order three, i.e., . Then, there is no quotient group of isomorphic to : the only normal subgroups of are the trivial subgroup, , and , and in none of these cases is the quotient of order three.

### Example of non-abelian groups of order eight

`Further information: dihedral group:D8, subgroup structure of dihedral group:D8, quaternion group, subgroup structure of quaternion group`

We can take as the dihedral group of order eight and as the cyclic maximal subgroup of dihedral group:D8, which is isomorphic to cyclic group:Z4.

Alternatively, we can take as the quaternion group and as one of the cyclic maximal subgroups of quaternion group, which is isomorphic to cyclic group:Z4.

### Example of arbitrary degree symmetric groups

If we take to be the symmetric group of degree , , and to be *any* subgroup other than the trivial subgroup, the whole subgroup, and a cyclic subgroup of order two, then is not isomorphic to any quotient of .