Cyclic Group Proof

Date: 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!