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October 7th, 2007, 07:15 PM   #1
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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.
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October 7th, 2007, 09:17 PM   #2
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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).
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August 14th, 2008, 12:03 PM   #3
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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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