Subgroup need not be isomorphic to any quotient group
Contents
Statement
It is possible to have a group (in fact, we can choose
to be a finite group) and a Subgroup (?)
of
such that there is no normal subgroup
of
for which
is isomorphic to the Quotient group (?)
.
Related facts
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
Proof
Example of symmetric group of degree three
Further information: symmetric group:S3, subgroup structure of symmetric group:S3, A3 in S3
Let be the symmetric group of degree three (order six) acting on the set
and
be the unique subgroup of order three, i.e.,
. Then, there is no quotient group of
isomorphic to
: the only normal subgroups of
are the trivial subgroup,
, and
, and in none of these cases is the quotient of order three.
Example of non-abelian groups of order eight
Further information: dihedral group:D8, subgroup structure of dihedral group:D8, quaternion group, subgroup structure of quaternion group
We can take as the dihedral group of order eight and
as the cyclic maximal subgroup of dihedral group:D8, which is isomorphic to cyclic group:Z4.
Alternatively, we can take as the quaternion group and
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 to be the symmetric group of degree
,
, and
to be any subgroup other than the trivial subgroup, the whole subgroup, and a cyclic subgroup of order two, then
is not isomorphic to any quotient of
.