Usbrth

Is it true that the order of the group $U(n)$ for $n>2$ is always an even number? If yes, how to go about proving it? U(n) is the set of positive integers less than n and co-prime to n ,which is a group under multiplication modulo.

The order of the group $U(n)$ is given by Euler's totient function $phi(n).$

If $n=2^m$ then $phi(n)=2^m-2^m-1,$ which is even for $m>1 ;(i.e., n>2)$.

Otherwise $n$ has a factor of the form $p^m$ with $p$ an odd prime and $mge1$,

in which case $phi(n)$ is a multiple of $p^m-p^m-1$,

so $phi(n)$ is even since $p^m-p^m-1$ is (being the difference of two odd numbers).

The order of the cyclic group $U(n)$ is $phi(n)$, as there are $phi(n)$ elements $ < n$ which are relatively prime with $n$(these are preciely the elements which possess an inverse in the cyclic group). And $phi(n)$ addreses even values after $n geq 2$

If $x$ is in your group $U(n)$, then so is $n-x$, because any common divisor of $n$ and $n-x$ would also divide their difference $x$. So $U(n)$ has exactly as many elements $<frac n2$ as it has $>frac n2$. Since $frac n2$ itself isn't in $U(n)$ (being a nontrivial divisor of $n$ --- this is where you use that $n>2$), the number of elements in $U(n)$ is double the number of elements of $U(n)$ that are $<frac n2$. In particular, it's an even number.

In case you've already learned Lagrange's theorem, here's an alternative proof. $U(n)$ has an element of order $2$, namely $n-1$. By Lagrange, the order of any element in a group divides the order of the group. So $2$ divides the order of $U(n)$.