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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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i wanted to get some hints how to get further and a feedback if this is a good start. –  André Oct 13 '12 at 11:27
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Suppose $|Z(G)| = p$. Show that if $g$ is an element of $G$ such that $g \not\in Z(G)$, we have $C_G(g) = G$, a contradiction. Alternatively, you could use the fact that $G/Z(G)$ cyclic implies that $G$ is abelian. –  Mikko Korhonen Oct 13 '12 at 11:30
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Also, it does not really make sense to say that "each coset is cyclic". If $H$ is a subgroup, then the only coset of $H$ that is a group is $H$ itself, the others do not contain the identity since different cosets are disjoint. –  Mikko Korhonen Oct 13 '12 at 11:31
    
Ok i have done it as follows: Assume $|Z(G)| = p$ and $g \notin Z(G)$. Then $\neg(\forall x \in G: xg = gx)$ so that we have a $x \in G$ with $xg \neq gx$. Then we have $x \notin C_G(g)$ and so $|C_G(g)| < p^2$. But then we must have $|C_G(g)| = p$ because $|C_G(g)| > 1$. With $Z(G) \subseteq C_G(g)$ we have $C_G(g) = Z(G)$. But then must $g \in Z(G)$ which is a contradiction. –  André Oct 13 '12 at 11:55
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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. –  Mikko Korhonen Oct 13 '12 at 12:09
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marked as duplicate by Gerry Myerson, rschwieb, Mikko Korhonen, Martin Sleziak, Thomas Oct 13 '12 at 14:17

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3 Answers

Cosets aren't necessarily subgroups, so they can't really be "cyclic". That definition applies to groups.

There are groups with a cyclic center that are non-abelian, for example any group with a trivial center. So I doubt you can conclude simply from the fact that $Z(G)$ is cyclic. Here is an hint though: suppose $G \neq Z(G)$ (ie. $G$ is nonabelian). Consider an element $x \in G$, $x \not\in Z(G)$. What can you say about the normalizer $N_x = \{y \in G : xy=yx \}$?

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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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In fact you only have $Z(G)=p$ or $Z(G)=p^2$. So $\big|G/Z(G)\big|=1$ or $\big|G/Z(G)\big|=p$. In the first one you find your group abelian and in the second one $aZ(G)$ for any $a\in G$ can generate $G/Z(G)$ so again you have $G=\langle a, Z(G)\rangle$. –  B. S. Oct 13 '12 at 11:54
    
@BabakSorouh Well, it ends up that the second case is impossible, and you have $G=Z(G)$ in both cases. –  rschwieb Oct 14 '12 at 0:51
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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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