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I have the following problem:

Let G be a group of order $p^2$ and assume that G is not abelian. I have just shown that $Z(G)$ is cyclic and has order p. Now I want to show that for each element $g\in G\setminus Z(G)$ we have $\operatorname{ord}(g)=p$.

Proof

Let $g\in G\setminus Z(G)$. We remark that $$card(G\setminus Z(G))=card(G)-card(Z(G))=p^2-p=p(p-1)$$Since we know that $\operatorname{ord}(g)\mid card(G)=p(p-1)$ we get that $\operatorname{ord}(g)=\{p,p-1,1\}$. Now let us assume that $\operatorname{ord}(g)=p-1$. Then $g^{p-1}=e$. But this means that $$g^p=g^{p-1}g=eg=g \Rightarrow p=1$$ Which is a contradiction to the assumption that p is prime. Now let us assume that $\operatorname{ord}(g)=1$. Then for all $g\in G\setminus Z(G)$ we have $g^1=g=e$. This means that $G\setminus Z(G)={e}$ which is also a contradiction. Therefore $\operatorname{ord}(g)=p$.

Does this works like this?

José Carlos Santos
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user123234
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  • yes sure but this exercise will lead me to a contradiction so that we prove at the end that every group of order $p^2$ is abelian. But does my proof works like this? – user123234 Oct 30 '21 at 13:51
  • The possible orders are $1,p,p^2$. If there is an element of order $p^2$, the group is cyclic, hence abelian. Thus, if your group is declared to be non-abelian, the possible orders are $1$ and $p$. – lulu Oct 30 '21 at 13:53
  • @Strandkorb ah, fair enough. – Randall Oct 30 '21 at 13:54
  • You don't have $\mathrm{ord}(g) \mid \mathrm{card}(G \setminus Z(G))$ since $G \setminus Z(G)$ is not a group. – aschepler Oct 30 '21 at 13:55
  • @aschepler ah good point! – user123234 Oct 30 '21 at 13:56

1 Answers1

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The solution is simpler than this. First I'd like to point out (one reason) why your proof is incorrect - it says that $\text{card}(G \setminus Z(G)) = p(p-1)$, and then uses the fact that $\text{card}(G) = p(p-1)$. The order of $G$ is $p^2$.

This logic can still be used. Since the order of $g$ divides $\text{card}(G)=p^2$, we know that the order must be contained in $\{1,p,p^2\}$. If the order is 1, the element is the identity. If the order is $p^2$, then the group is abelian, because it is cyclic. Therefore the order of any non identity element must be $p$. In particular, any element not in the center must have order $p$.

Nerif
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