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Suppose that $G$ is a group with more than one element, $G$ had no proper, non-trivial subgroup then prove that $|G|$ is prime.

Attempt.

Claim

$G$ is finite

If not then for any $x\neq e$ we have $\langle x^2 \rangle$ a non-trivial subgroup of $G$. Hence $G$ is finite.

Now given $G$ is finite.Let $|G|=m$ For any $x\neq e$ we have $\langle x \rangle$ a subgroup of $G$.

Now because there exists no non trivial subgroup, we have $\langle x\rangle=G$

Hence $G=\langle x \rangle$

Hence $G$ is cyclic.

How do I show that $|G|$ is prime?

Kindly do not use Cauchy Theorem. Use Lagrange's Theorem only, or topics taught before Lagrange Theorem.

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  • $\begingroup$ What if $x^2=e?$ That's possible when $G$ is infinite. It's even possible for it to be true for all $x\in G.$ $\endgroup$ – Thomas Andrews Jun 14 '18 at 2:51
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    $\begingroup$ @ThomasAndrews then you would have a subgroup of order $2$, therefore non-trivial and proper ($\left<x\right>$). $\endgroup$ – Arnaud Mortier Jun 14 '18 at 2:51
  • $\begingroup$ Sure, but that's not what the argument said. @ArnaudMortier $\endgroup$ – Thomas Andrews Jun 14 '18 at 2:52
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    $\begingroup$ @ArnaudMortier A person learning math shouldn't skip steps. An important step here is stating for any $x\neq e,$ $\langle x\rangle =G.$ So $\langle x\rangle$ is infinite, and hence $\langle x^2\rangle$ is a proper subgroup. $\endgroup$ – Thomas Andrews Jun 14 '18 at 2:55
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    $\begingroup$ @user567182 Sure, but your proof that $G$ is finite is incomplete. And $\langle x \rangle = G$ is a useful statement for both the finite and infinite cases. $\endgroup$ – aschepler Jun 14 '18 at 2:59
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For any $x\in G$, $x\neq e$ we must have $\langle x\rangle=G$, since $H=\langle x\rangle$ is a subgroup, and $H\neq \{e\}$.

Now, if $x^2=e$, we have $G=\langle x\rangle=\{e,x\},$ a cyclic group of order $2$.

If $x^2\neq e$ then $\langle x^2\rangle=G$, too, so $x=(x^2)^k$ for some $k,$ and hence $x^{2k-1}=e$ for some $k.$ In particular, $G=\langle x\rangle$ is finite.

If $x$ is of order $n$ where $n$ is not prime, say $n=ab,$ with $a,b>1.$ Then $\langle x^a\rangle\neq G.$

So, we must have $n$ prime, and we are done.

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Here is the usual complete argument.

Let $x\in G$, with $x \ne e$. Then $\langle x \rangle$ is a nontrivial subgroup and so must be $G$, that is, $G=\langle x \rangle$.

If $G=\langle x \rangle$ is infinite, then $\langle x^2 \rangle$ is a proper subgroup. Therefore, $G$ is finite.

Thus, $G=\langle x \rangle$ is cyclic of order $m$. If $m=ab$, with $a,b>1$, then $\langle x^a \rangle$ is a proper, non-trivial subgroup. Therefore, $m$ is prime.

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  • $\begingroup$ This assumes $G$ is finite. $\endgroup$ – Thomas Andrews Jun 14 '18 at 3:05
  • $\begingroup$ @ThomasAndrews, my answer was a continuation of the OP's argument. I've added the complete argument. $\endgroup$ – lhf Jun 14 '18 at 11:11

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