TITLE: If $n > 2$, prove that the order of the multiplicative group of units modulo n, $U_n$, is even.
QUESTION [5 upvotes]: I'm struggling with this. I know it is going to use Lagrange's Theorem.  This is what I have so far.
Suppose $|U_n| = k.$ 
This implies $a^k = 1$ for all $a$ in $U_n$ and $|a|$ divides $k$.
Now, what can be said about the order of an element $a$?
I haven't been able to conclude that it's even with what I have, therefore I continued so:
$|\langle a\rangle|$ divides $k$ where $\langle a\rangle$ is the cyclic subgroup generated by $a.$
However, I haven't been able to conclude that $|\langle a\rangle|$ is even. 
So I tried using this:  an element $a$ is in $U_n$ iff gcd$(a, n) = 1$.
Any hints?

REPLY [4 votes]: Yet another hint: Show that $\{+1,-1\}$ is a subgroup of $U_n$ of order $2$. Now apply Lagrange.