Thanks, hope your exam went well.Quote:

Sorry I don't have much time to write it out myself for you, I have to get to a final exam.

basically p divides |G| and each term of the sum, but if |Z(G)|=1 this would be impossible (move the sum over to the other side and p divides the left side, but not the right. Z(G) is a subgroup, so by lagrange its order has to divide $\displaystyle p^2$ and we see it is not 1.

Here is a proof of the G/Z(G) cyclic thing.

http://crypto.stanford.edu/pbc/notes...quotient.xhtml

Sorry again for the brevity.