I'm trying to provide a solution to the following claim:
"The order of $U(n)$ is even when $n>2$."
Note: here, $U(n)$ is the set of all positive elements that are less than and relatively prime to $n$.
I think I've correctly proven the claim, but I couldn't find anything similar online. So, I would appreciate any criticism/corrections regarding my proof (me being fairly new to abstract algebra, help is very welcome!).
proof: Assume $|U(n)|=m$, for $n>2$. Firstly, suppose that some $\mu \in U(n)$ has an even order. By Lagrange's Theorem, we know that $|\mu|$ divides $m$. Thus, $m$ is even (since for $k\in \mathbb{N}$, $\not \exists l \in \mathbb{N} : 2k|(2l+1)$). If $\mu$ is odd, then $m$ is either even or odd. However, we know (I forgot the theorem's name) that the number of elements of order $2$ is divisible by $\phi(2)=1$. So, there is at least one element of order $2$ in $U(n)$. Thus, if $m$ is odd then $\exists h \in U(n)$ such that $ |h|$$\not |$$m$, which is impossible. Therefore, $m=|U(n)|$ must always be even. $\blacksquare$ (?)
Best Answer
I like your idea that if $U(n)$ has an element of even order, then the order of $U(n)$ is even by Lagrange's Theorem. On the other hand, for $n>2$, the order of $n-1$ in $U(n)$ is 2.
Another approach to this problem is to work with properties of the Euler phi function since $o(U(n))=\varphi(n)$.