2.59

It is possible that G has a normal subgroup N but has no subgroup isomorphic to G/N. Actually one can take G to be the group of integers Z and N its subgroup of even integers 2Z. N is normal in G because G is abelian, and G/N = Z/2Z is cyclic of order two. But G = Z has no subgroup that is cyclic of order two, since no element of Z has order two. This is the most illustrative example.

For counter-example with G finite, one cannot use cyclic groups, since a quotient of a finite cyclic group G is cyclic, and G has cyclic subgroups of every possible order. The symmetric group S_n has only one normal subgroup if n is equal to three or at least five, it has index two, and S_n does have elements of order two. S₄ has a normal subgroupf index six, with quotient isomorphic to S₃, but S₄ does have subgroups isomorphic to S₃. The dihedral group D_n has a normal subgroup of index two, and cyclic subgroups of order two, and, when n is even, a normal subgroup of index n (namely {1,-1}) and quotient D_m, 2m = n, but also has a subgroup isomorphic to D_m.

There is another non-abelian group of order 8, the quaternion group Q₈ = {1,-1,i,-i,j,-j,k,-k}, with i, j, and k multiplying like as in the vector cross product, and i² = j² = k² = -1. This group G has a unique subgroup N of order two {1,-1}, which is central, hence normal, and every element of G/N has order two. Then G/N is isomorphic to Z₂ x Z₂ and G has no such subgroup (since it doesn't have three elements of ordet two). (Thanks to Ryan for pointing out this example.)