Why can't be ($\mathbb{Z}_5,+) $ isomorphic to $U(R)$ for some rings $R$ where $U(R)$ is the unity groups of $R$? Why can't  be  $(\mathbb{Z}_5,+) $  isomorphic  to $U(R)$  for  some  rings $R$  where $U(R)$  is  the  unity  groups  of  $R$  ? Is important  that $(\mathbb{Z}_5,+)$ is  cyclic?
 A: Suppose $R$ is a ring with $U(R) = \langle u \rangle$ and $u^5 =1$.
We must have $1+1=0$ or else $-1$ would be an element of order $2$ in $U(R)$.
Now consider the subring of $R$ generated by $u$.
Its invertible elements must be the same as $R$'s (if someone is invertible in the subring, it is invertible in $R$), and it is the image of $\Bbb F_2[U]/(U^5+1)$ in $R$, so it is enough to check all the rings that are quotients of $\Bbb F_2[U]/(U^5+1)$.
Now, by the Chinese Remainder Theorem, $\Bbb F_2[U]/(U^5+1) \cong \Bbb F_2[U]/(U+1) \times \Bbb F_2[U]/(1+U+U^2+U^3+U^4) \cong \Bbb F_2 \times \Bbb F_{16}$.
Its only quotient rings are the zero ring, $\Bbb F_2$, $\Bbb F_{16}$, and $\Bbb F_2 \times \Bbb F_{16}$ itself. Their unit group are either trivial or cyclic of order $15$.

This can be turned into a simple concrete proof. $1+u$ has order $15$ mod $1+u+u^2+u^3+u^4$, and so we should look at $v = (u^2+u^3+u^4)$ (this is $1$ mod $1+u$ and $1+u$ mod $1+u+u^2+u^3+u^4$) :
$v = u^2(1+u+u^2)$
$v^2= u^4(1+u^2+u^4) = u(1+u^2+u^3)$
$v^3 = u^3((1+u+u^2)+(u^2+u^3+u^4)+(u^3+u^4+1)) = u^4$
$v^5 = (1+u^2+u^3)$
$v^{15} = (u^4)^5 = 1$
Therefore, $v$ is invertible ($v.v^{14} = 1$), so by assumption, we must have $v^5 =1$. This entails $u^2 = u^3$ and then $u=1$, contradiction. 
