Question: Prove that every group of order p^2 (where p is a prime) is abelian.

 

What I started to say was this:

|G| = p^2, then |G/Z(G)| = |G|/|Z(G)| which can equal p^2, p or 1, since those are the divisors of p. If |G/Z(G)| = p, then G/Z(G) is cycic because one element has order 1 and the rest of order p, which means there is a generator, it is cyclic and then G is abelian.

If |G/Z(G)| = 1, then G=Z(G) so it much be abelian. However, the case where |G/Z(G)|=p^2 is where I get a little hung up. I know it never happens, but how can i prove it?

It's the class equation again - the centre can never be trivial in a p-group.

How do we know that the center can never be trivial? The center can be trivial when the order of the group is qp, where q and p are primes. I am confused here, would p have to be in the center if the order of G were p^2?

how can p be in the centre? p is a number?

 

Look up the class equation: the order of the group is the sum of the sizes of the conjugacy classes. All conjugacy classes have order a power of p, and there is at least one conjugacy class with a single element in it, so....