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If $G$ has no non trivial subgroups, then Show that $G$ must be of prime order. This question is from Herstein Page 46 Question 3.

Attempt:

Let $G$ has prime order(say $p$). By Lagrange theorem, order of the subgroup can be $1$ or $p$. So it has two trivial subgroups.

How do i prove the other way around?

Thanks.

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    $\begingroup$ "Let G has prime order(say p) .So by Lagrange theorem ,order of subgroup can be 1 or p .So it has two non trivial subgroups" These $are$ the trivial subgroups $\endgroup$ – Jack Mar 25 '15 at 3:38
  • $\begingroup$ @Jack sorry i intended to write no non trivial $\endgroup$ – Taylor Ted Mar 25 '15 at 3:39
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    $\begingroup$ The result is not quite true, the group whose only element is $e$ has no non-trivial subgroups, and $1$ is not prime. Now suppose $G$ has an $a\ne e$. The subgroup $A$ generated by $a$ is cyclic, and because $G$ has no non-trivial subgroups, this group is all of $G$, so $G$ is cyclic. If $G$ is infinite cyclic, you can easily produce a non-trivial subgroup. Now take care of finite cyclics with composite order. $\endgroup$ – André Nicolas Mar 25 '15 at 3:42
  • $\begingroup$ @AndréNicolas i understand what you have written ,How should i go about my proof $\endgroup$ – Taylor Ted Mar 25 '15 at 3:47
  • $\begingroup$ Only one item left to do. Let $n=ab$ where neither $a$ nor $b$ is equal to $1$. Show that $\mathbb{Z}_n$ has a non-trivial subgroup. $\endgroup$ – André Nicolas Mar 25 '15 at 3:49
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Consider the subgroups generated by one element - that is, those of them form $\{e,x,x^2,x^3,\ldots\}$. Now, if $G$ is to have no non-trivial subgroups, then if we choose some $x$ other than the identity, this subgroup cannot be the trivial group - thus, for it to be a trivial subgroup, it's got to be the whole group $G$ (which, as an aside, means $G$ is a cyclic group). Now, to finish, consider: if $|G|=ab$, what can we say about the subgroup generated by $x^a$?

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  • $\begingroup$ And it is not infinite. $\endgroup$ – Daniel Valenzuela Mar 25 '15 at 5:19
  • $\begingroup$ @Dan True, though if you let $|G|=ab$ be cardinal arithmetic, with $a$ finite, it works fine in the infinite case. $\endgroup$ – Milo Brandt Mar 25 '15 at 21:52

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