The order of the group $G$, meet the following conditions: $1
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
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group-theory
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0What do you mean by a neutral number? Since this looks like a homework problem, what have you tried? – 2012-07-16
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0The order of G is not infinite. – 2012-07-16
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1Note 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? – 2012-07-16
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0@Lag: Those are called **natural** numbers, not "neutral" numbers. – 2012-07-16
1 Answers
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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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0Actually, Cauchy should not be needed for this question. – 2012-07-16
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0@Tobias: Fair enough, though it obviates the need to prove that a group with a particular property must have prime order, no? – 2012-07-16
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0I am not sure what you mean exactly. I just felt that though Cauchy can certainly be used, it seems like overkill in this case. – 2012-07-16
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0@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 – 2012-07-16
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0If G=H1 then by using legrange, the order of G must be prime. But what if $G\not=H1\not=H2$ – 2012-07-16
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0@Lag since the condition must hold for all subgroups, there is no need to consider that case. – 2012-07-16
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0OK, so lets go back. G = H1. Then $gcd(|G|,|H2|)=1$ Using lagrange, and the hint from above, $0<|H2|<|G|$ and $|H2| | |G|$, then $gcd(|G|,|H2|)=|H2|$ so |H2| must be 1 and |G| can be everything.... – 2012-07-16
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0What am I doing wrong in that solution? – 2012-07-16
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0@Lag: You haven't given a solution, you've just made a statement. So, technically, there is nothing wrong in "that solution", because there is no such solution. You've concluded that $G$ has the property that every proper subgroup is trivial. Keep going. – 2012-07-16
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0We know that if the order of G is prime, then the group cycle. Let say the the order is NOT prime. Then if $G=H1\not=H2$ we get that $gcd(|G|,|H2|)=1$ from the conditions. By using lagrange, we get that there is a sub group |H2|=prime and, $|G|/|H2| = n$ where $n\not=1$ (because if n=1 then |G| is prime). But if n is have to be 1, because of the condition. Therefor |G| is prime, and G cycle. – 2012-07-16
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0@Lag: No, Lagrange doesn't tell you $|H_2|$ is prime, it tells you that $|H_2|=1$. The question is: why is a group whose only proper subgroup is the trivial subgroup cyclic? – 2012-07-16
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0What I meant was, by lagrange there is a subgroup |H2| that |G|/|H2|=n. If |G| is not prime, then there is a subgroup that |H2|=prime – 2012-07-16
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0@ArturoMagidin I understand how you got to ask that question. But I cant find any proper way to explain it. Maybe it is something to do that H2 is the maximum subgroup of G? – 2012-07-16
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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