(1) Prove that if |G|=1365, then G is not simple.
(2) Assume that G is a nonabelian group of order 15. Prove that Z(G)=1. Use the fact that the group generated by "g" is less than or equal to C_G(g) for all "g" in G to show that there is at most one possible class equation for G.
Simple Groups and Sylow p-Subgroups are investigated.