October 7th, 2007, 08:15 PM  #1 
Newbie Joined: Oct 2007 Posts: 7 Thanks: 0  no subgroups
Let G be a group which has no subgroups except itself and {e}. Prove that G is either {e} or a cyclic group of prime order.

October 7th, 2007, 10:17 PM  #2 
Site Founder Joined: Nov 2006 From: France Posts: 824 Thanks: 7 
By Cauchy theorem, if the order n of G were not prime, then your group would have a cyclic subgroup of order p where p divides n (alternatively, you can use the more elaborate Sylow theorem).

August 14th, 2008, 01:03 PM  #3 
Member Joined: Aug 2008 Posts: 84 Thanks: 0  Re: no subgroups
You don't need Cauchy's or Sylow's theorem(s) for this (this problem is in the chapter on cyclic groups and subgroups in Herstein's Topics). The first step is take a g in G, where g is not the identity element. Then the set <g> is a subgroup of G, but since G has no nontrivial subgroups we must have G = <g>, so G is cyclic. The prime order follows as soon as you prove that it is finite.


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