Prove that $\mathbb{Z}_n$ has an even number of generator for n>2 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.
 A: 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.
A: I like it.  Alternatively, the number of generators of $\Bbb Z_n$ is $\varphi(n)$, where $\varphi$ is Euler's totient function.  A basic fact about $\varphi$ is that it's values are always even.
For your second question, the hint is to look at cyclic subgroups, noting that the infinite cyclic subgroup, $\Bbb Z$, has infinitely many subgroups.
A: Let $\mathbb{Z}_n$ be cyclic group of order $n \gt 2$ then either $n = 2^s$ such that $s \gt 1$, say if $s = 1$ then $n = 2^1 = 2$ but $n \gt 2$ so $s$ has to be more $1$.
Then no. of generators of $\mathbb{Z}_n$ = no. of elements of order $n=$
$$ \varphi(n) = \varphi(2^s) = 2^s - 2^{s-1} = 2^{s-1}(2-1) = 2^{s-1}$$
which is divisible by $2$ for $s$ more than 1, so $\varphi(n)$ is even.
If $n$ is not of the form $2^s$ then $n$ must be of the form $p^kt$ such that $p$ is prime greater than $2$, $p$ and $t$ are co-prime and $n$ is $t$ multiple of $p^k$ then
(Can you guess further steps of the proof?)

 Spoiler
 $\varphi(n) = \varphi(p^kt)$ $= \varphi(p^k)* \varphi(t)$ $= (p^k - p^{k-1})* \varphi(t)$ $= p^{k-1}(p-1)* \varphi(t)$. Note that $p$ is prime more than $2$ then $p$ must be odd and $p-1$ must be even and as $\varphi(n)$ is multiple of $(p-1)$ then $\varphi(n)$ must be even.


Hence $\mathbb{Z}_n$ must have even no. of generators if $n$ is more than 2. Infact you can generalize above result for any cyclic group with order more than $2$.
