Let \(p\) be any prime number.
Let \(G\) be a group with order \(p^2\). We show that \(G\) is abelian.
As \(G\) has order of the form \(p^n,\) we know that \(Z(G) \neq (1).\) (See proof here.)
Thus, \(|Z(G)| = p\) or \(|Z(G)| = p^2.\) In the latter case, we’re done as \(Z(G) = G \implies G\) is abelian.
Assume \(|Z(G)| = p.\) We shall arrive at a contradiction.
Choose \(x \in G\setminus Z(G).\) (This set is nonempty as \(Z(G)\) has order strictly less than that of \(G.\))
Consider \(H = Z(x) = \{g \in G : gx = xg\},\) the set of all elements in \(G\) that commute with \(x.\)
It is easy to check that \(H\) is a subgroup of \(G.\)
Claim 1: \(Z(G) \lneq H\)
Proof. The (improper) containment follows immediately by definition.
To see that it’s proper, observe that \(x \in Z(x)\) but \(x \notin Z(G),\) by our choice of \(x.\)
Claim 2: \(H \lneq G\)
Proof. Once again, the proper containment is immediate, by definition of \(H.\)
To see the proper containment, suppose \(H = G.\) In that case, \(x\) commutes with every element of \(G.\) However, then we get that \(x \in Z(G),\) a contradiction.
Thus, by the above two claims, we get that \(Z(G) \lneq H \lneq G.\)
This leads to a contradiction as we get that the order of \(H\) is strictly between \(p\) and \(p^2\) but also divides \(p^2.\)
To summarise, we get that the only possibility is that \(Z(G) = G\) which gives us that the groups is abelian as every element commutes with every other.