Prove cyclic group with one generator can have atmost 2 elements

Question

Prove cyclic group with one generator can have atmost 2 elements .

Attempt

Consider a cyclic group generated by $a \neq e$ ie G = .So G is also generated by <$a^{-1}$> .Now Since it is given that there is one generator thus $a = a^{-1}$ which implies that $a^{2}=1$ .Using $a^{O(G)}=e$ .$O(G)=2 $

But i am not confident with this

Thanks

Answer 1

Consider an arbitrary element $g \in G$. Since $G=\langle a \rangle$, then there is $k \in \mathbb{Z}$ such that $g=a^k=a^{2q+r}$, where $r \in \{0,1\}$ is the remainder of the euclidean division of $k$ by $2$, and $q$ is the quotient. Then $g=(a^2)^q \cdot a^r = 1^q \cdot a^r = a^r$ for some $r \in\{0,1\}$. So $g \in \{1, a\}$ which has at most two elements.

Answer 2

Since $a^2 = e$, either $a$ is the identity or $a$ has order $2$. If $a$ is the identity, $G = \{e\}$; if $a$ has order $2$, $G = \{e,a\}$.