Did you try anything or have any ideas yet? How far did you get?
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.