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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EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not endomorphic kernel|View examples of subgroups satisfying property normal subgroup and endomorphic kernel
It is possible to have a group (in fact, we can choose to be a finite group), and a normal subgroup of such that there is no subgroup of isomorphic to the quotient group . In particular, need not be an endomorphism kernel in .
- 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.
Example of the quaternion group
Suppose is the quaternion group of order eight, and is the center of quaternion group . Then is a Klein four-group. However, all the subgroups of order four in are isomorphic to cyclic group:Z4 and hence, in particular, has no subgroup isomorphic to .