Suppose G is a finite group and I know for every k≤|G| that exactly nk elements in G have order k. Do I know what the group is? Is there a counterexample where two groups G and H have the same number of elements for each order, but G is

notisomorphic to H? I suspect that there is, but I haven’t thought of one.

**Answer**

Take G=Z/4×Z/4, and H=Q8×Z/2 of order 16, where Q8 denotes the quaternion group. Both groups have exactly 1 element of order 1, 3 elements of order 2 and 12 elements of order 4.

Edit: I understood the question as follows: Is there a counterexample where two groups G and H have the same number of elements *for* each order, but G is not isomorphic to H ? Is it really required, that all elements different from 1 in G have the same order ?

**Attribution***Source : Link , Question Author : Stanley , Answer Author : Hooked*