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This question is from cyclic group exercise. I have read the theorems of cyclic groups. But I could mot answer with proper language. Help me to solve.

user409382
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2 Answers2

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Hint: Suppose that the order $n$ of $G$ is not prime, write $n=\Pi p_i^{n_i}$ where $p_i$ is a prime, $n_i\neq 0$. Sylow implies the existence of a subgroup of order $p_i^{l_i}$ $0\leq l_i\leq n_i$.

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Pick a non-identity element $g\in G$. If it has infinite order then the subgroup generated by $g^2$ is a proper subgroup. If it has finite order and it is not a prime you can pick a prime $p$ that divides $|g|$ and notice the group generated by $g^p$ is a proper subgroup. So the order of $g$ must be a prime $p$, we must also have that the subgroup generated by $g$ is all of $G$, so the order of $G$ is $p$.

Asinomás
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