If $n > 2$, prove that the order of the multiplicative group of units modulo n, $U_n$, is even.

Question

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?

Answer 1

Hint: Use a pairing argument. If $x$ is a unit, then so is $-x$. And if $n\gt 2$ they are distinct.

Answer 2

Different hint: $|U_n|=\varphi(n)$, the Euler totient function. So you have to prove that for $n \gt 2$ this number is even.

Answer 3

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