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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?

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    The solution I was thinking about myself when I gave my hint was slightly different. When $g \not\in Z(G)$, we have $Z(G) \subseteq C_G(g)$ and $g \in C_G(g)$. Thus |C_G(g)| > p which implies $|C_G(g)|= p^2$ and $C_G(g) = G$. Therefore $g \in Z(G)$, a contradiction.2012-10-13

2 Answers 2

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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.

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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.)

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    @BabakSorouh Well, it ends up that the second case is impossible, and you have $G=Z(G)$ in *both* cases.2012-10-14