If a cyclic group $C=\langle c\rangle$ has an involution $z$, then
- $C$ is finite
- $C$ has even order
- the involution is unique
If $C$ was not finite, then it must be isomorphic to $(\mathbb Z,+)$ which has no element $a\neq0$ s.t. $a+a=2a=0$, hence $C$ must be finite.
Then $C$ is isomorphic to $\mathbb Z/n\mathbb Z$ for some $n$. And here $2a=0$ is true iff $n$ is even.
Finally I noticed that an involution must be $z=c^{\frac n2}$, so it's necessarely unique.
Am I right? Thank you all.
I'll assume that, by involution in a (multiplicative) group $G$, you mean an element $g\in G$ such that $g^2=1$ and $g\ne 1$.
You're correct in your conclusions, but I would show the details in a better form.
Lagrange's theorem implies that a finite group with an involution has even order. So, in your case, $C$ must be isomorphic to $\mathbb{Z}/(2k)\mathbb{Z}$ for some $k>0$.
Cauchy's theorem, on the other hand, implies that any finite group with even order has at least an involution.
Now, passing to additive notation, your involution will be an element $z+2k\mathbb{Z}$ with $0<z<2k$ and $$ 2z\in 2k\mathbb{Z}. $$ This means $z\in k\mathbb{Z}$, so $z=ky$. Then from $0<ky<2k$ you conclude that $0<y<2$, which leaves little to imagination.
;-)