Show that no $U$-group of order $16$ is isomorphic to $\mathbb{Z}_4\oplus \mathbb{Z}_4$ $$U(n)=\{x : 0<x<n, \gcd(x,n)=1\}.$$
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.
 A: 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.)
A: Lemma1: $Z_n\cong Z_r\times Z_s$ if and only if $(r,s)=1$ where $n=rs$. 
Lemma2:if $G\cong N\times K$ and $  (|N|,|K|)=1 $ then $Aut(G)\cong Aut(N)\times Aut(K) $ .
When $n$ is prime $U(n)$ is cylic so we can assume that $n$ is not prime.Thus we can write $n=rs$ s.t. $(r,s)=1$ if it has more than one prime.
By lemma1, $Z_n\cong Z_r\times Z_s$ and by lemma2, $U(n)\cong Aut(Z_r)\times Aut(Z_s)$ so if $U(n)\cong Z_4\times Z_4$ then $$Aut(Z_r)\cong Aut(Z_s)\cong Z_4$$
Then it is easy to see that $Z_r\cong Z_s\cong Z_5\implies$ $Z_n\cong Z_5\times Z_5$ but it is not cyclic $\implies contradiction.$
If it has only one prime divisior,i.e $n=p^k$ then $\phi(n)=p^k-p^{k-1}=16 \implies p\in \{2,17\}$ .Since we know that $p$ can not be $17$,then $p=2$ and $k=5$.
But $Aut(Z_{32})$ can not be isomorphic to $Z_4\times Z_4$ since order of $3$ in $Z^*_{32}$ is more than $4$.$3,9,27,17,19...$ in mod $32$. We are done.
For  lemma2 please see proposition 5.3 and 5.4  in http://math.uchicago.edu/~may/REU2013/REUPapers/Sommer-Simpson.pdf
A: Case 1: $n=pq$ for distinct odd primes $p$ and $q$
$U(n)\cong U(p)\oplus U(q)\cong \mathbb Z_{p-1}\oplus\mathbb Z_{q-1}$
In this both can never be simultaneously 4, as this will imply $p=q$, contrary to the choice of $p$ and $q$. 
Case 2: $n= 2^{k}$, $k>1$
$U(n)\cong\mathbb Z_2\oplus\mathbb Z_{2^{k-2}}$ which is again not isomorphic to $\mathbb Z_4\oplus\mathbb Z_4$ 
Case 3: $n= p.2^{m}$, $p$ is odd prime, $m>1$
$ U(n)\cong U(p)\oplus U(2^{m})\cong\mathbb Z_{p-1}\oplus\mathbb Z_2\oplus\mathbb Z_{2^{m-2}}$. 
Hence, we proved for any possible value of n which satisfies $ |U(n)|=16 $
