$$U(n)=\{x : 0<x<n, \gcd(x,n)=1\}.$$
We are asked to show that no $U$-group of order $16$ is isomorphic to $\mathbb{Z}_4\oplus \mathbb{Z}_4$. (External direct product)
I started calculating various possibilities for n.
I used a technique taught in number theory that the order clause is satisfied by
n=17, 32, 34, 40, 48, 60. But i'm not allowed to use this result. I have no clue how to proceed.
Any help would be appreciated.
Best Answer
Take the prime power factorization of $n$ as $n=\prod_j p_j^{a_j}$. Then we have a ring isomorphism
$\mathbf{Z}/n\mathbf{Z} \cong \prod_j \mathbf{Z}/p_j^{a_j}\mathbf{Z}$ ( by Chinese Remainder Theorem). Then taking the unit groups on either sides we get $U(n) \cong\prod_j U(p_j^{a_j})$. Your requirement states that there be just 2 terms, cyclic of order 4,on the rhs. One can work out now that it is not possible for any $n$. (Perhaps you can dismiss the case of $n$ being twice an odd number first.)