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.
Let G be a non trivial group with no non trivial proper subgroup. Prove that G is a group of prime order.
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cyclic-groups
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1Have you heard of Lagrange's theorem about finite groups? – 2017-01-24
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2Yes sir, I have heard lagranges theorem. – 2017-01-24
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1Well then, assume $H $ is a nontrivial subgroup of $G$. What does that imply? – 2017-01-24
2 Answers
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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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1$|G|=\infty$ must also be taken into consideration. – 2017-01-24
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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 $p$.