Cyclic Group ProofDate: 12/15/2008 at 19:44:19 From: Steven Subject: Cyclic groups How can I prove that if a group G has a unique subgroup of order d for every d that divides |G|, then G is cyclic? I can prove the converse but this gives me problems. Date: 12/17/2008 at 03:25:48 From: Doctor Jacques Subject: Re: Cyclic groups Hi Steven, The general idea of the proof is that a cyclic group has the smallest possible number of cyclic subgroups. Let us write phi(d) for the number of generators of a cyclic group of order d (actually, phi(d) is Euler's totient function, but we do not need to know that). We also write phi(1) = 1. Let G be a group of order n. If C is a cyclic subgroup of order d of G, let us write gen(C) for the set of generators of C--there are phi(d) of them. Now, every element of G generates a unique cyclic subgroup, which means that G is the disjoint union of gen(C) when C ranges over the _cyclic_ subgroups of C. This means that we have, in general: |G| = n = SUM (c(d)*phi(d)) [1] where c(d) is the number of cyclic subgroups of order d, and d ranges over the divisors of n. Note that, if G happens to be cyclic of order n, all the c(d) are equal to 1, as you have shown, and we have: n = SUM (phi(d)) [2] Note that [2] is a purely number-theoretic identity--it only depends on n, not on the structure of G. If we know that there is exactly one subgroup of order d for each d, then there is at most one _cyclic_ subgroup of order d, and we have c(d) <= 1 for all d in [1]. If we compare [1] and [2], we see that we must have c(d) = 1 exactly for all d; in particular, we must have c(n) = 1, which means that there is a cyclic subgroup of order n, which is G itself. Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 12/18/2008 at 00:05:58 From: Steven Subject: Thank you (Cyclic groups) That's great, thanks! |
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