# Does every nontrivial finite cyclic group have prime order?

Here's my attempt as a proof. Did I make any mistakes?

Let $G \ne \langle e \rangle$ be a finite cyclic group. Then any $a \in G \setminus \{e\}$ has order $|G|$. Aiming for a contradiction, suppose $|G| = xy$ for some integers $x,y \ge 2$. Then $e = a^{|G|} = a^{xy} = (a^x)^y$. Since $x < |G|$, we know $a^x \ne e$, thus $a^x\in G \setminus \{e\}$ has order less than $|G|$ (specifically, its order is a divisor of $y$, which is a divisor of $|G|$). This contradicts that every $a \in G \setminus \{e\}$ has order $|G|$.

Therefore, any nontrivial finite cyclic group must have prime order.

If it's correct, is there anything I can do to improve its clarity?

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The second sentence is false. The smallest counterexample is the finite cyclic group $\mathbb{Z}/4\mathbb{Z}$ of order $4$, which has an element of order $2$. –  Qiaochu Yuan Mar 27 '13 at 2:55
Where does your first claim ('Then any $a\in G$ has order $|G|$') come from? What finite cyclic groups do you know, and have you tried plugging in some examples to this proof? –  Steven Stadnicki Mar 27 '13 at 2:56
And BTW: the claim is very not true...lest you'd think your supposed proof is flawed but there's other one correct. –  DonAntonio Mar 27 '13 at 2:59

The statement you claim to have contradicted, i.e. that every element of a cyclic group $G$ has order either $1$ or $|G|$, is false. For instance, you can convince yourself that for any integer $n$, the set $\{0,1,2,\dots,n-1\}$ endowed with addition modulo $n$ is a cyclic group generated by $1$. This provides many counterexamples. In fact, for any composite $n$ we have a cyclic group of nonprime order.
"Then any $a \in G \backslash {e}$ has order $|G|$."
Not technically true. The actual statement is that for any $a \in G$, the order of $a$ divides the order of $G$. This, of course, isn't really at all special for cyclic groups - it's just a special case of Lagrange's theorem, where the two items happen to both be generated by only one element. In any case, it's clear that for any $\mathbb{Z}/n\mathbb{Z}$, any number that's not relatively prime to $n$ won't have equal order.