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Let $G$ be a non-trivial cyclic group with an unique generator.

Show that $G$ has exactly two elements.


Let $g$ be a generator of $G$. My first idea to approach the problem was to show that if $|G|=n > 2$ then there is an integer $k$ between 1 and $n$ such that $\langle g^k \rangle = G$ or, equivalently, $g \in \langle g^k \rangle$.

My second idea was to analyse the problem in 2 separate cases. For the first one, if $n$ is prime, let $k$ be a positive integer then, by Lagrange's Theorem, $|g^k| \mid n$ and, therefore, $\langle g^k \rangle = n$. Hence, we must conclude that, if $n > 2$ the generator cannot be unique.

For the second case, where $n$ is composite, I was hoping to get to a contradiction, but could not advance any further. Is there a way to develop either of these ideas or is it necessary a new approach?

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Hint: show that if $G=\langle g\rangle$ then we also have $G=\langle g^{-1}\rangle$. If G has a unique generator, then that element must be its own inverse, so the group has order 2.

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Greicius answer is formidable, but your first instinct was correct. I would like to expound your idea.

If $G=<a>$ is infinite, then $a \in <a^{-1}> $ and $G$ has a no unique generator, so $G$ must be finite.

Now, if $G$ is finite, observe that if $G = <a>$ is finite of order $n$, then $a^k$ is a generator of $G$ if and only if $n$ and $k$ are relatively prime. In particular, the number of generators of $G$ is then $\phi (n)$, i.e. $\vert\{k\in\mathbb{N} \vert 1\leq k < n \land (n,k)=1\}\vert$.

Now this last statement needs to be proven.

If $n,k\in\mathbb{N}$ and $(n,k)=1$, let $H=\{(a^k)^n\vert n\in\mathbb{N}\}$. Clearly, $H \subset G$. Then there are integers $p,q \in \mathbb{Z}$ s.t. $pn+qk=1$, hence $a=a^1=a^{pn+qk}=(a^{n})^p (a^{k})^{q}=(a^{k})^{q}$, since $a^n=1$. So the powers of $a^k$ generate $a$, but $<a>$ is the smallest subgroup containing the element $a$, so $G\subset H$ and hence $G=H$, i.e. $a^k$ generates $G$.

Conversely, let $G=<a^k>$, then $\exists q\in\mathbb{Z}$ s.t. $(a^{k})^{q}=a$. That means $kq \equiv 1\!\!\!\mod n$, or - equivalently - $\exists m\in\mathbb{Z}$ s.t. $kq-1=mn$ and hence $kq-mn=1$. So $n$ and $k$ must be relatively prime.

Having proved this, there is only to observe that $\phi(n)\geq 2$ for $n \geq 3$.

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