Let $G=\langle a\rangle$ be a cyclic group. If $G$ is infinite, then $a$ and $a^{-1}$ are the only generators of $G$. If $G$ is finite of order $m$, then $a^k$ is a generator of $G$ if and only if $(k,m) = 1$.
My attempt: Suppose $G= \langle a\rangle$ is infinite. It’s easy to check, $G=\langle a^{-1}\rangle$. Assume towards contradiction, $\exists k\in \Bbb{Z}\setminus \{1,-1\}$ such that $G= \langle a^{k}\rangle$. If $k=0$, then $\langle a^{0}\rangle = \langle e\rangle =\{e\}$. Which contradicts our initial hypothesis of $G$ is infinite. If $|k|\gt 1$. Since $G$ is infinite, we have $a^i\neq a^j$, if $i\neq j$. So $a\notin \langle a^{k}\rangle$ and $G\neq \langle a^{k}\rangle$. Thus we reach contradiction. Hence $G=\langle a^k\rangle$ if and only if $k=1$ or $k=-1$.
Claim: If $a\in G$ with $|a|=m$ and $k\in \Bbb{N}$, then $\langle a^k\rangle = \langle a^{(k,m)}\rangle$. Here $G$ is an arbitrary group. Proof: Let $(k,m)=d$. Let $x\in \langle a^k\rangle$. Then $x=a^{ki}$ for some $i\in \Bbb{Z}$. Since $d|k$, we have $k=dj$ for some $j\in \Bbb{Z}$. So $x=a^{ki}=(a^{d})^{ij}\in \langle a^d\rangle$. Thus $\langle a^k\rangle \subseteq \langle a^{d}\rangle$. Let $x\in \langle a^d\rangle$. Then $x=a^{di}$ for some $i\in \Bbb{Z}$. By Bezout identity, $\exists r,s\in \Bbb{Z}$ such that $d=kr+ms$. Since $a^m=e$, we have $$x=a^{di}=a^{kri+msi}=(a^k)^{ri}(a^m)^{si}=(a^k)^{ri}e= (a^k)^{ri}.$$ So $x=(a^k)^{ri}\in \langle a^k\rangle$. Thus $\langle a^k\rangle \supseteq \langle a^d\rangle$. Hence $\langle a^k\rangle =\langle a^d\rangle$.
Suppose $G=\langle a\rangle$ and $|G|=m\in \Bbb{N}$. By definition, $|a|=|G|=m$. $(\Leftarrow)$ Let $(k,m)=1$. By above claim, $\langle a^k\rangle =\langle a^{(k,m)}\rangle =\langle a\rangle$. Thus $G=\langle a^k\rangle$. $(\Rightarrow )$ Let $\langle a\rangle = \langle a^k\rangle$, for some $k\in \Bbb{N}$. By above claim, $\langle a\rangle = \langle a^{(k,m)}\rangle$. Assume towards contradiction $(k,m)=d\geq 2$. By theorem 4(vii) section 1.3, we have $$| \langle a\rangle|= | \langle a^d\rangle|=|a|=|a^d|=m=\frac{m}{d}\lt m.$$ Thus we reach contradiction. Hence $(k,m)=1$. Is my proof correct?
