Let be the representatives of the conjugacy classes of the finite group G and assume that these elements pairwise commute. Prove that G is abelian.
Hmm.. well that's not the way to go. Just because representatives of classes commute, doesn't mean that all elements between the classes commute. For example, consider the classes . There must be another way..
EDIT: Try this.
First, we need the fact that, for any finite group and subgroup , .
Now, we let act on itself by conjugation. Since it is assumed that the commute pairwise, we have for all . That is, for each , (the stabilizer of under this group action).
Let be arbitrary. Then there exist such that . Hence for each .
We have shown that, for any , we can find so that . It follows that (for each ).
By the claim above, since is finite, this forces for each . That is, the conjugacy class of is . We know that this means and, since these elements form a complete set of representatives for the conjugacy classes of , , and the group is abelian.