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It is given that the group $G$ has only two subgroups: the trivial group $\{e\}$ and $G$ itself, can we be sure that $G$ has order $k$ which must be a prime number?

I know that the reverse of this statement is true by Lagrange's theorem, but cannot be sure whether the above statement is true or not.

user26857
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Rescy_
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  • Have you seen Cauchy's theorem? – Thomas Credeur Nov 07 '15 at 00:17
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    Hints: First establish that $G$ is cyclic. Next establish that $G$ must have finite order. Then conclude that $|G|$ must be prime. – Geoff Robinson Nov 07 '15 at 00:19
  • @ThomasCredeur I have not learnt Cauchy's theorem, though I know Lagrange's theorem. My textbook 'Algebra and Geometry' by A.F.Beardon does not cover Cauchy's theorem. – Rescy_ Nov 07 '15 at 00:19
  • @GeoffRobinson I am struggling with the proof of finiteness. If $G$=${e}$ then the proof is over. Otherwise let $g \in G$, then $G$ is the group generated by $g$. However I feel difficult to prove finiteness. – Rescy_ Nov 07 '15 at 00:21
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    Suppose that $g$ above does not have finite order. Consider the subgroup $\langle g^{2} \rangle$. Show that it is neither the identity subgroup, nor all of $G$. – Geoff Robinson Nov 07 '15 at 00:25
  • @ThomasCredeur : It does not seem to be given that the group is finite. – Geoff Robinson Nov 07 '15 at 00:29
  • @Rescy_ $G={e}$ seems to be a counterexample, unless you consider $1$ to be a prime number. – bof Feb 08 '16 at 09:01

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Take a non trivial element $x$ of the group. Then the subgroup generated by this element $x$ is $\langle x\rangle=G$. Note that $G$ cannot be infinite because otherwise $G$ will be isomorphic to $\Bbb{Z}$, and it's known that all subgroups are of the forms $n\Bbb{Z}$ (so an infinite numbers of subgroups). Now to prove that the cardinal is prime, suppose $n$ is not prime, that is, $n=rs$ for some integers $\ge 2$. What can you said about the order of $x^s$?

user26857
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JeSuis
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