Quotient group need not be isomorphic to any subgroup
From Groupprops
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)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about endomorphic kernel
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
Contents
Statement
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
.
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
Proof
Example of the quaternion group
Further information: quaternion group, subgroup structure of quaternion group, center of 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
.