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The order of the group $G$, meet the following conditions: $1 where n is a natural number.

For each 2 sub groups $H_1$, $H_2$ of $G$, if $H_1 \neq H_2$ then $\gcd(|H_1|,|H_2|)=1$. (gcd = greatest common divisor)

Prove that the order of $G$ is a prime number and the group is cycle.

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    @Lag: Those are called **natural** numbers, not "neutral" numbers.2012-07-16

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Hint 1. If $0\lt a\leq b$ and $a|b$, then $\gcd(a,b) = a$.

Hint 2. Lagrange and Cauchy are your friends.

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    @Lag: If you know Cauchy's Theorem, then it follows from Cauchy's Theorem. If you don't know Cauchy's Theorem, then there is a direct proof that doesn't use much, but I'm not going to just give it to you on a silver platter. I'll simply say: pick $x\in G$, $x\neq e$. Go from there.2012-07-17