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I know that a group of order 5 does not have any nontrivial factors and that you cannot factor 5 as a product of two numbers larger than 1 but I do not know where to go from here?

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HINT: If $a$ is an element of order $5$, what is the order of $a^2$?

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Hint: Prove that for any g, $o(g) = o(g^{-1})$.

The statement will remain true when 5 is replaced by any $n \geq 3$.

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Sorry, I was confused. – Graphth Sep 26 '12 at 23:57

If $x$ has order $5$, it means, $x^5=1$ in the group. So, $x^2$ will also have order 5, but $x^2\ne x$ because then $x=1$ would follow by division.

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