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) need not satisfy the second subgroup property (i.e., endomorphic kernel)
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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 $G$.

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

• Subgroup need not be isomorphic to any quotient group

Proof

Example of the quaternion group

Further information: quaternion group, subgroup structure of quaternion group, center of 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 $K$.