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Let $G$ be a group with order $p$ a prime number. Show $G$ is cyclic.

I know that if $G$ is a cyclic group and if $|G|=n$, then $G\cong\mathbb{Z}_n$. But I don't where to go from there.

Any help/hints would be welcome. ^_^

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marked as duplicate by Zev Chonoles, JSchlather, Marc van Leeuwen, Stahl, Micah Sep 16 '13 at 7:03

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The definition of a cyclic group is a group such that all the elements of the group generated by one element. One can also show that if $G$ is cyclic of order $n$ then it is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ (I'm sure you can find this fact in your book). Hence, you probably want to find an element that has the order of the group to finish your problem. Here is a hint:

Let $x$ be an element of $G$ that is not the identity. What does Lagrange theorem tell you? That the order of $x$ divides the order of $G$, which is a prime. Which are the divisors of a prime?

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I chose this answer because I was looking for a hint, not an entire solution. Thank you. –  Desperate Fluffy Sep 16 '13 at 6:39
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Suppose $\forall a \in G,a\neq1$, because the order of element is the factor of $p$, so we get $$a^p=1$$

that means that order of $a$ is an factor of $p$,and $a\neq1$ so $Order_a=p$. so $$G=\langle a \rangle $$ a cyclic group

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Please read this meta thread about using quantifiers correctly, and read this meta thread about using MathJax correctly (< and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle). –  Zev Chonoles Sep 16 '13 at 7:00
thanks i have corrected as you've said –  Alex Sep 16 '13 at 7:05
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