# Prove that G is cyclic if distinct subgroups have coprime orders

The order of the group $G$, meet the following conditions: $1<G<n$ 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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What do you mean by a neutral number? Since this looks like a homework problem, what have you tried? –  Tobias Kildetoft Jul 16 '12 at 12:17
The order of G is not infinite. –  Lag Jul 16 '12 at 12:19
Note that the condition you mention also has to hold when $H_1 = G$ and $H_2$ is any proper subgroup of $G$. What does Lagrange then tell you? –  Tobias Kildetoft Jul 16 '12 at 12:22
@Lag: Those are called natural numbers, not "neutral" numbers. –  Arturo Magidin Jul 16 '12 at 12:57

Hint 1. If $0\lt a\leq b$ and $a|b$, then $\gcd(a,b) = a$.
@Tobias: After Lagrange, you can conclude that all proper subgroups of $G$ are don't want to give it away. It is a standard exercise in beginning group theory to prove "A group in which every proper subgroup is that thing must be cyclic of prime order. This is, indeed, usually proven before you prove Cauchy's Theorem, but it takes a couple of lines. Cauchy gives it to you in 1 line if you already know that $G$ is finite –  Arturo Magidin Jul 16 '12 at 13:12
If G=H1 then by using legrange, the order of G must be prime. But what if $G\not=H1\not=H2$ –  Lag Jul 16 '12 at 13:18