Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
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
2  
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
add comment

2 Answers

up vote 1 down vote accepted

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.

share|improve this answer
add comment

"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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.