Subgroup need not be isomorphic to any quotient group

Statement

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

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

• Quotient group need not be isomorphic to any subgroup

Proof

Example of symmetric group of degree three

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

Let $G$ be the symmetric group of degree three (order six) acting on the set $\{ 1,2,3 \}$and $H$ be the unique subgroup of order three, i.e., $H = \{ (), (1,2,3), (1,3,2) \}$. Then, there is no quotient group of $G$ isomorphic to $H$: the only normal subgroups of $G$ are the trivial subgroup, $H$, and $G$, 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 $G$ as the dihedral group of order eight and $H$ as the cyclic maximal subgroup of dihedral group:D8, which is isomorphic to cyclic group:Z4.

Alternatively, we can take $G$ as the quaternion group and $H$ 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 $G$ to be the symmetric group of degree $n$, $n \ge 3$, and $H$ to be any subgroup other than the trivial subgroup, the whole subgroup, and a cyclic subgroup of order two, then $H$ is not isomorphic to any quotient of $G$.