proof that every group of prime order is cyclic

The following is a proof that every group of prime order is cyclic.

Let $p$ be a prime and $G$ be a group such that $|G|=p$ . Then $G$ contains more than one element. Let $g\\in G$ such that $g\eq e_{G}$ . Then $\\langle g\\rangle$ contains more than one element. Since $\\langle g\\rangle\\leq G$ , by Lagrange’s theorem, $|\\langle g\\rangle|$ divides $p$ . Since $|\\langle g\\rangle|>1$ and $|\\langle g\\rangle|$ divides a prime, $|\\langle g\\rangle|=p=|G|$ . Hence, $\\langle g\\rangle=G$ . It follows that $G$ is cyclic.

单词	ProofThatEveryGroupOfPrimeOrderIsCyclic
释义	proof that every group of prime order is cyclic The following is a proof that every group of prime order is cyclic. Let $p$ be a prime and $G$ be a group such that $\|G\|=p$ . Then $G$ contains more than one element. Let $g\\in G$ such that $g\eq e_{G}$ . Then $\\langle g\\rangle$ contains more than one element. Since $\\langle g\\rangle\\leq G$ , by Lagrange’s theorem, $\|\\langle g\\rangle\|$ divides $p$ . Since $\|\\langle g\\rangle\|>1$ and $\|\\langle g\\rangle\|$ divides a prime, $\|\\langle g\\rangle\|=p=\|G\|$ . Hence, $\\langle g\\rangle=G$ . It follows that $G$ is cyclic.
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