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Prove that a group with more than one element contains an element of prime order.

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    True for "finite" groups.2017-01-10
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    Presumably you mean a finite group with more than one element? And where are you stuck?2017-01-10

2 Answers 2

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So your group has a non identity element $a$. Since the group is finite, the order of $a$ is finite. Say $|a| = p_1^{n_1}\dotsb p_k^{n_k}$ where the $p_i$ are prime. What is the order of the element $$\large a^{\;p_1^{n_1}\dotsb p_k^{n_k-1}}\;?$$

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pick a non identity element $g$, suppose $p$ divides $|g|$, then $g^{|g|/p}$ has order $p$.