Prove that a group with more than one element contains an element of prime order.
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True for "finite" groups. – Anurag A Jan 10 '17 at 06:04
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Presumably you mean a finite group with more than one element? And where are you stuck? – carmichael561 Jan 10 '17 at 06:04
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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 $$\large a^{\;p_1^{n_1}\dotsb p_k^{n_k-1}}\;?$$
Mike Pierce
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pick a non identity element $g$, suppose $p$ divides $|g|$, then $g^{|g|/p}$ has order $p$.
Asinomás
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