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 fact about::subgroup $$H$$ of $$G$$ such that there is no normal subgroup $$N$$ of $$G$$ for which $$H$$ is isomorphic to the fact about::quotient group $$G/N$$.

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

Example of symmetric group of degree three
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
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 $$G$$.