How Can G/Z(G) Theorem Prove a Group Abelian When |G|=p^2?

[tyrannosaurus]

Homework Statement

(1)To prove this I have to let G be a group, with |G|=p^2.
(2)Use the G/Z(G) theorem to show G must be Abelian.
(3) Use the Fundamental Theorem of Finite Abelian Groups to find all the possible isomorphism types for G.

Homework Equations

Z(G) = the center of G (a is an element of G such that ax=xa for all x in G)

The Attempt at a Solution

I can prove it by using Conjugacy classes and gettting that the order of Z(G) must be non-trival and going on from there, but we have not gotten to Conjugacy Classes yet so i can't use this fact. Can anyone help me on this? I know that |Z(G)|=1 or pq when the order of |G|=pq where p and q are not distinct primes. From there I am unsure on how to uise the G/Z(G) thereom to prove that G is abelian.

 

[mighty2000]

Thank you for your post. You are correct in thinking that the G/Z(G) theorem can be used to prove that G is abelian. Here is a potential solution:

(1) Let G be a group with |G|=p^2, where p is a prime number.
(2) By the G/Z(G) theorem, we know that G/Z(G) is isomorphic to a subgroup of Aut(G), where Aut(G) is the group of automorphisms of G.
(3) Since |G|=p^2, Aut(G) has order p^2(p-1). This means that G/Z(G) must have order 1 or p.
(4) If G/Z(G) has order 1, then G is abelian, since Z(G) is the entire group in this case.
(5) If G/Z(G) has order p, then G/Z(G) is cyclic, since all subgroups of a cyclic group are also cyclic.
(6) By the Fundamental Theorem of Finite Abelian Groups, we know that a cyclic group of order p has only one isomorphism type, which is isomorphic to Zp.
(7) Therefore, G/Z(G) is isomorphic to Zp, which means that G is abelian.
(8) By the Fundamental Theorem of Finite Abelian Groups again, we know that G is isomorphic to either Zp^2 or Zp x Zp.
(9) Thus, the possible isomorphism types for G are Zp^2 or Zp x Zp.

I hope this helps. Let me know if you have any further questions or if you would like me to clarify any steps. Good luck with your proof!