Is U-group isomorphic to $\mathbb{Z}_4 \oplus\mathbb{Z}_4$? How can i show that no $U(n)$ is isomorphic to $\mathbb{Z}_4 \oplus\mathbb{Z}_4$? Please give me some hint. Thanks in advance.
 A: I take $U(n)$ to mean the group of invertible elements in $\Bbb{Z}_{n}$.
If
$$
n = \prod p_{i}^{e_{i}}
$$
is the decomposition on $n$ as the product of  powers of distinct primes, then
$$
U(n) \cong \oplus U(p_{i}^{e_{i}}).
$$
So we ask first (thanks @AlexYoucis) whether we can have
$$
U(p_{i}^{e_{i}}) \cong \Bbb{Z}_4 \times \Bbb{Z}_4.
$$
By order reasons, this could only happen for $p_{i}^{e_{i}} = 17$ (but then $U(p_{i}^{e_{i}}) \cong \Bbb{Z}_{16}$ is cyclic of order $16$), or for $p_{i}^{e_{i}} = 32$, as $\varphi(32) = 16$. But $U(2^k) \cong \Bbb{Z}_{2^{k-2}} \times C_2$ is known to be non-cyclic for $k \ge 3$.
Then we ask when we can have
$$
U(p_{i}^{e_{i}}) \cong \Bbb{Z}_4.
$$
This can only happen when $p_{i}^{e_{i}} = 5$, because $U(8) \cong \Bbb{Z}_2 \times \Bbb{Z}_2$. And it is easy to see that $\varphi(n) = 4$ only for these two values.
So, indeed no $U(n)$ is isomorphic to $\Bbb{Z}_4 \times \Bbb{Z}_4$.
A: Here is a slightly silly approach. Assume $U(n)$ is isomorphic to $\mathbb{Z}_4\oplus \mathbb{Z}_4$. This forces $\phi(n)=16$. One can play around with this and the formula 
$$\phi(p_1^{r_1}\cdots p_s^{r_s})=(p_1^{r_1}-p_1^{r_1-1})\cdots (p_s^{r_s}-p_s^{r_s-1})=p_1^{r_1-1}\cdots p_s^{r_s-1}(p_1-1)\cdots (p_s-1)$$ 
to find that the only such $n$'s are 
$$
17,\;32,\;34, \;40,\;48,\; 60.
$$
By the chinese remainder theorem $U(k\cdot l)\simeq U(k)\oplus U(l)$ whenever $k$ and $l$ are relatively prime. For $p$ prime $U(p)\simeq \mathbb{Z}_{p-1}$. Besides, due to the presence of $-1$ (order $2$) and $5$ (order $2^{r-2}$) whose generated subgroups have trivial intersection,
$$
U(2^r)\simeq \mathbb{Z}_2\oplus \mathbb{Z}_{2^{r-2}}.
$$
It follows that $17$ and $34=2\cdot 17$ yield $\mathbb{Z}_{16}$. Impossible.
Then, $40=2^3\cdot 5$, $48=2^3\cdot 3$, and $60=2^2\cdot 3\cdot 5$ yield three nontrivial $\mathbb{Z}_{2^j}$ summands. Still impossible.
And finally $32=2^5$ gives $\mathbb{Z}_2\oplus \mathbb{Z}_8$. It does not work either.
