October 7th, 2007, 07:15 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, 09:17 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, 12:03 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.


Tags 
subgroups 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Subgroups  gaussrelatz  Algebra  1  October 10th, 2012 11:30 PM 
Normal subgroups  Fernando  Abstract Algebra  1  April 11th, 2012 12:40 AM 
Number of subgroups of Sn  honzik  Abstract Algebra  0  February 13th, 2011 06:45 AM 
Normal Subgroups  ejote  Abstract Algebra  4  February 1st, 2011 07:30 AM 
Subgroups  DanielThrice  Abstract Algebra  1  November 25th, 2010 02:28 PM 