Subgroup need not be isomorphic to any quotient group

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It is possible to have a group G (in fact, we can choose G to be a finite group) and a Subgroup (?) H of G such that there is no normal subgroup N of G for which H is isomorphic to the Quotient group (?) G/N.

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Example of symmetric group of degree three

Further information: symmetric group:S3, subgroup structure of symmetric group:S3, A3 in S3

Let G be the symmetric group of degree three (order six) acting on the set \{ 1,2,3 \}and H be the unique subgroup of order three, i.e., H = \{ (), (1,2,3), (1,3,2) \}. Then, there is no quotient group of G isomorphic to H: the only normal subgroups of G are the trivial subgroup, H, and G, 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 G as the dihedral group of order eight and H as the cyclic maximal subgroup of dihedral group:D8, which is isomorphic to cyclic group:Z4.

Alternatively, we can take G as the quaternion group and H 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 G to be the symmetric group of degree n, n \ge 3, and H to be any subgroup other than the trivial subgroup, the whole subgroup, and a cyclic subgroup of order two, then H is not isomorphic to any quotient of G.