Let H be a team of stimulate 3. By meaning of group, there deserve to be just one identity element in the team H.

You are watching: Prove that a group of order 3 must be cyclic

So, $H=\left \ e,x,y \right \$.

By definition of cyclic group,

we have actually that the aspects x and y

$x=g^n \exists n \in \urbanbreathnyc.combbZ$

$y=g^n \exists n \in \urbanbreathnyc.combbZ$

In particular, n is confident for if it were not, a contradiction would arise from having an ext than one identification element.

Any clues or assistance is appreciated.

Thank in advance.

abstract-algebra group-theory
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asked Jun 2 "16 at 10:15

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$\begingroup$ In third and 4th line of her attempt of systems you it seems to be ~ to assume the thesis ("$H$ is cyclic") as a hypothesis, making your discussion circular. $\endgroup$
Jun 2 "16 in ~ 10:19

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2 answers 2

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You"re trying to PROVE that $G$ is cyclic, for this reason you can not (yet) assert that $x = g^n$. Instead, think about $x \cdot x$. It have to be either $x, y,$ or $e$. If it"s $x$, then you have$$x^2 = x\\x^2 (x^-1) = x x^-1\\x = e$$which is a contradiction, because $x$ and also $e$ are distinct aspects of the group.

If $x^2 = e$, then $x$ has order 2, yet 2 does no divide 3, for this reason this contradicts Lagrange"s theorem.

Finally, we conclude that $x^2 = y$, and thus the team is cyclic, produced by the facet $g = x$.

Alternative if friend don"t choose Lagrange yet:

In the instance where we intend that $x^2 = e$:

The aspects $xe, xx,$ and $xy$ must all be distinctive for if two were the same, climate multiplying by $x^-1$ top top the left would display that 2 of $e, x, y$ to be the same, i m sorry is impossible.

Since $xe = x$ and we"re presume $x^2 = e$, us must have $$xy = y.$$multiplying ~ above the best by $y^-1$ offers $x = e$, a contradiction. For this reason $x^2 = e$ is likewise impossible.

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answered Jun 2 "16 at 10:20

john HughesJohn Hughes
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Actually, any kind of group of prime order is cyclic.

Indeed, allow $p=\lvert G\rvert$ and $a\in G$, $a\ne e$. By Lagrange"s theorem, the subgroup $\langle\mkern1.5mu a\mkern1.5mu \rangle$ produced by $a$ has order a divisor that $\lvert G\rvert=p$, and is $>1$ since $a\ne e$, whence $\lvert\langle\mkern1.5mu a\mkern1.5mu \rangle\rvert=p$, i.e. $\;\langle\mkern1.5mu a\mkern1.5mu \rangle=G$.

reply Jun 2 "16 in ~ 10:35

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