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.
Prove that G is cyclic if distinct subgroups have coprime orders
1
$\begingroup$
group-theory
-
0@Lag: Those are called **natural** numbers, not "neutral" numbers. – 2012-07-16
1 Answers
2
Hint 1. If $0\lt a\leq b$ and $a|b$, then $\gcd(a,b) = a$.
Hint 2. Lagrange and Cauchy are your friends.
-
0@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