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.
Best Answer
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).