Show that any cyclic group of even order has exactly one element of order $2$ Show that any cyclic group of even order has exactly one element of order $2$


Attempt:

Let $G$ be a finite group of even order.

Assuming a set $A=\lbrace g \in G \vert g \neq g^{-1} \rbrace$, I can show that  $A\cup\{e\}$ has odd number of elements as $A$ has even number of elements. So $G$ has at least one element $a$ such that $a$ does not in $A$, ie $a=a^{-1}$ ie $a^2=e$. That is $G$ has atleast one element of order $2$. "More specifically, $G$ has odd number of elements of order $2$ --- ???"

But how to show that exactly one such element if $G$ be cyclic.


  NB: Please don't use isomorphic properties.
If $G$ is a finite cyclic group of order $n$, then $G$ has a unique subgroup of order $m$ for each $m$ dividing $n$. Now, if $n$ is even, then $G$ has a unique subgroup of order $2$, hence one element of order $2$ (as if $G$ had two distinct elements of order $2$, then we'd have two distinct subgroups of order 2).
Let $g$ be a generator of $G$ and $G$ has order $n$.  Suppose, $(g^m)^k=e$. Then $km=nr$ for some integer $r$. Let $d=(m,n)$ and $l=mn/(m,n)$.  Note that $l$ is the least common multiple of $n$ and $m,n$ and $km=nr=lt$ for some $0\le t&lt;d$. Now, $k=n/d$ and $o(g^k)=d$. Letting $m=2$, we get that $o(g^{m/2})$ is the only element of order $2$.