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Question: Prove that the group of order 3 is cyclic.

Attempt:

Let H be a group of order 3. By definition of group, there can be only one identity element in the group H.

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

By definition of cyclic group,

we have that the elements x and y

$x=g^{n} \exists n \in \mathbb{Z}$

$y=g^{n} \exists n \in \mathbb{Z}$

In particular, n is positive for if it were not, a contradiction would arise from having more than one identity element.

Any hints or assistance is appreciated.

Thank in advance.

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  • $\begingroup$ In third and fourth line of your attempt of solution you seem to assume the thesis ("$H$ is cyclic") as a hypothesis, making your argument circular. $\endgroup$ – user228113 Jun 2 '16 at 10:19
  • $\begingroup$ What is $g$ here? $\endgroup$ – Servaes Jun 2 '16 at 10:19
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You're trying to PROVE that $G$ is cyclic, so you cannot (yet) assert that $x = g^n$. Instead, consider $x \cdot x$. It must 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 $e$ are distinct elements of the group.

If $x^2 = e$, then $x$ has order 2, but 2 does not divide 3, so this contradicts Lagrange's theorem.

Finally, we conclude that $x^2 = y$, and thus the group is cyclic, generated by the element $g = x$.

{Alternative if you don't like Lagrange yet:}

In the case where we suppose that $x^2 = e$:

The elements $xe, xx,$ and $xy$ must all be distinct for if two were the same, then multiplying by $x^{-1}$ on the left would show that two of $e, x, y$ were the same, which is impossible.

Since $xe = x$ and we're assuming $x^2 = e$, we must have $$ xy = y. $$ multiplying on the right by $y^{-1}$ gives $x = e$, a contradiction. So $x^2 = e$ is also impossible.

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Actually, any group of prime order is cyclic.

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

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  • $\begingroup$ elegant proof!. $\endgroup$ – Cloud JR Aug 12 '18 at 11:53

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