Hi All,
How to I Claim: For any interger n> 1, the group u(n(n+2)) is not cyclic?
Thanks.
If $\displaystyle \gcd(n,m) = 1$ then $\displaystyle \text{U}(\mathbb{Z}/nm\mathbb{Z}) \simeq \text{U}(\mathbb{Z}/n\mathbb{Z}) \times \text{U}(\mathbb{Z}/m\mathbb{Z})$ by the Chinese remainder theorem.
Now if $\displaystyle n$ is odd then $\displaystyle \gcd(n,n+2) =1$ and so $\displaystyle \text{U}(\mathbb{Z}/n(n+2)\mathbb{Z}) \simeq \text{U}(\mathbb{Z}/n\mathbb{Z}) \times \text{U}(\mathbb{Z}/(n+2)\mathbb{Z})$. Now this cannot be cyclic because the orders of those two groups are not relatively prime, since $\displaystyle \phi(x)$ is for $\displaystyle x>1$.
Similarly with $\displaystyle n$ odd, just write, $\displaystyle n(n+2) = 4(n/2)(n/2+1)$ and apply the argument.