My proof of this question is as following
Let $x$ be the element of the generator for the cyclic group $\mathbb{Z}_n$, so that that the order of $x$ is $n$, i.e. $x^n=x^{n-1}\circ x=x\circ x^{n-1}=e$, so $x^{n-1}$ must be the inverse of $x$ in the group. Then because $\gcd(n,n-1)=1$, then the order of $x^{n-1}$ is $n/1=n$. Therefore $x^{n-1}$ is a generator of the group $\mathbb{Z}_n$, since $<x>=<x^{n-1}>$. Hence for every element $x$ that is the generator of the group $\mathbb{Z}_n$, we can always have its inverse element $x^{n-1}$ or $x^{-1}$ as another generator of the group. i.e. if I find odd numbers of generator for the group, I must have an even number of generators, or if I find even number of generators, I must have an even number of generators as well.
I don't know if my proof is correct enough so I want to see what your guys thinking. And perhaps if it is possible, I have a second question which I don't know how exactly to prove. here is the statement
A group with a finite number of subgroups is finite.
I intuitively suggest this must be true but I can't write reasonable proof for it.
Thanks.
Best Answer
More abstract: suppose that the group $G$ is cyclic; then, for each generator $x$, also $x^{-1}$ is a generator, because $x^k=(x^{-1})^{-k}$.
Suppose $x=x^{-1}$; then $x^2=1$ (or $e$, if you prefer this notation; I don't) and therefore $|G|\le2$.
Thus, if $|G|>2$, we have $x\ne x^{-1}$, for every generator $x$, and thus we can divide the generators into pairs.
Hint for the second question: any group is the union of its cyclic subgroups; if the number of subgroups is finite, none of the cyclic subgroups can be infinite.