Quotient group need not be isomorphic to any subgroup

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

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

 * Subgroup need not be isomorphic to any quotient group

Example of the quaternion group
Suppose $$G$$ is the quaternion group $$\{ 1,-1,i,-i,j,-j,k,-k \}$$of order eight, and $$H$$ is the center of quaternion group $$\{ 1, -1 \}$$. Then $$K = G/H$$ is a Klein four-group. However, all the subgorups of order four in $$G$$ are isomorphic to cyclic group:Z4 and hence, in particular, $$G$$ has no subgroup isomorphic to $$K$$.