3
$\begingroup$

Is my argument correct, if not what is wrong with it. If the group $G$ is not cyclic, this means that there is not element of order 3. So if $G=\{e, a, b\}$ then $|a|=2$ and $|b|=2$, a contradiction because by Lagrange's Theorem element order divide group order but 2 does not divide 3. Hence $G$ is cyclic.

I just want to show if this is correct or not and why, I do not need a proof.

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    Yep, correct as written.2012-11-04
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    @DonAntonio Thanks.2012-11-04
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    @Reader, have you had Cauchy's Theorem?: the order of an element must divide the order of the group. So if the group has order 3, then there must be an element of order 3 and hence powers of this element fill the whole group.2012-11-04
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    @ Nicky Hekster Thanks.2012-11-04
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    You do not need Lagrange's theorem to do the problem. Just the cancellation law is enough.2012-11-04
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    @Pantelis Damianou But the easiest argument is by using Lagrange. Am I right?2012-11-04
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    The most elementary method is to use Sudoku! The products $ab$ and $ba$ must be equal to $e$ (using cancellation law). By default $a^2=b$. Therefore $a^3=a^2 a =ba=e$.2012-11-04
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    @Pantelis Damianou Thanks2012-11-04

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