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I have a question about prime order group. This answer by amWhy says that:

It follows that any group of order 5 (and any group of prime order) must be generated by a single element and is hence, cyclic.

I understand previous part but this. If any group of prime order must be generated by a single element??

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eChung00
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Suppose $G$ is a group and that $G$ has prime order. Now pick some $g\in G$ with $g\not=e$. Then consider the group $\{g^n: n\in\mathbb{N}\}$. This is a nonzero subgroup of $G$. By LaGrange's theorem, its order must divide the order of $G$. But since the order of $G$ is prime, it must be all of $G$.

ncmathsadist
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    Either you spell it la Grange or Lagrange. The latter is what he used. The spelling LaGrange doesn't make sense. – egreg Apr 25 '15 at 17:52
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Let $g$ be a non-identity element of the group $G$. The order of $g$ is a divisor of the order of the group, which is prime, say $p$. The order isn't one, so it must be $p$.

Now in general, $|\langle g \rangle|$ equals the order of $g$. For us this means $|\langle g \rangle | = p$ which is the order of the group. So $\langle g \rangle$ is a subset of $G$ which has the same size as $G$. We must have $\langle g \rangle =G$, thus $G$ is generated by the element $g$.

Matthew Towers
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