Group with order $p^2$ must be abelian . How to prove that?

Question

Possible Duplicate:
Showing non-cyclic group with $p^2$ elements is Abelian

I must show that a group with order $p^2$ with $p$ prime must be a abelian. I know that $|Z(G)| > 1$ and so $|Z(G)| \in \{p,p^2\}$.

If I assume that the order is $p$ i get $|G / Z(G)| = p$ and so each coset of $Z(G)$ has order $p$ which means that each coset is cyclic and especially $Z(G)$ is cyclic. Can I conclude something by that?

Answer 1

Use the following theorem, probably the most important and basic in the theory of finite $\,p-\,$groups:

Theorem: The center of a finite $\,p-\,$group is non-trivial

Proof: Let $\,G\,$ be a finite $\,p-\,$group and make it act on itself by conjugation. Now just observe that:

$$(1)\;\;\;\;\;\;|\mathcal Orb(x)|=1\Longleftrightarrow x\in Z(G)$$

$$(2)\;\;\;\;\;\;\mathcal Orb(x)=[H:Stab(x)]\Longrightarrow p\mid|\mathcal Orb(x)|\;\;\;\;\;\square$$

Finally, the following lemma together with the above gives you what you want:

Lemma: For any group $\,G\,$ , $\,G/Z(G)\,$ is cyclic iff $\,G\,$ is abelian, or in otherwords: the quotient $\,G/Z(G)\,$ can never be non-trivial cyclic.

Answer 2

There is an easy exercise that if $Z$ is contained in $Z(G)$ and $G/Z$ is cyclic, then $G$ is abelian.

In the case of your problem, $G/Z(G)$ is cyclic, hence $G$ is abelian.

(The possible orders for $G/Z(G)$ are $p$ and $1$, but after applying the lemma, you find it to be 1.)