Thanks for Your reply! I just proved in previous problem that if p and q are distinct primes and G is non abelian then Z(G) is trivial but I did not notice the relation between these problems.

so quick proof:

From Lagrange Theorem |Z(G)| could be pq, p, q or 1. It is easy to prove that order of Z(G) can not be pq ( because in this case Z(G) = G therefore G is abelian what contradicts the initial claim ).

Also can not be p because the quotient group G/Z(G) would be cyclic of order q but G/Z(G) cannot be cyclic unless G is abelian ( because G/Z(G) cyclic implies G abelian.)

The same for q.

So Z(G) must be trivial.

And now I see that if |G| = 21 then |Z(G)| = 1 and from class equation I get that 3m+7n = 20 and must be n=m=2.

But I still can not done the 2nd question. I know that :

-the size of each orbit divides the group order

-in each conjugacy classes all elements have the same order

-in all abelian groups every conjugacy class is a set containing one element (if |G| =

and p is prime then G is abelian)

and I can infer that order of conjugacy classes of abelian group equals 1.

how to show that conjugacy classes in G have size p?