If $R$ is an integral domain, then $(R,+)$ is NOT isomorphic to $(U(R),*).$
That is, the additive group of an integral domain is not isomorphic to the group of units of the domain.
I tried to prove it, but I'm not very sure about the validity of my proof:
For the sake of argument, assume there is an isomorphism.
If $|R|=n, n \in \mathbb{N}$ then $|U(R)| \leq n-1$ because zero (identity element wrt addition) is not in the units since $0r=r0=0$ for every $r \in R.$
But this contradicts the fact that there is a bijection. Hence we conclude that there is no isomorphism in the case when $R$ is finite.
The following is false but I want to fix it and I don't know how.
Now assume $R$ is infinite, and suppose $\phi: (U(R),*) \rightarrow (R,+) $ is isomorphism.
Then $\phi(ab)= \phi(a)+\phi(b)$ and $\phi(1)=0$.
Therefore if $r \in (U(R),*),$ then so $r^{-1}$.
Let $r$ be arbitrary, then $\phi(1) = \phi(rr^{-1}) = \phi(r) + \phi(r^{-1}) = 0$
So $\phi(r) = - \phi(r^{-1}) \Rightarrow \phi(r) = \phi((r^{-1})^{-1}) $
Since $\phi$ is isomorphism, then $r=r^{-1}$.
But $r$ was arbitrary, and thence we get a contradiction.
I need to fix this
Another "proof" to the infinite case:
If the characteristic is zero:
$0 =\phi(1)=\phi(-1(-1))=\phi(-1)+\phi(-1)=2 \phi(-1) \Rightarrow \phi(-1)=0$ but as $\phi$ is 1-1 then $-1=1$, contradiction.
If the characteristic is not zero I don't know what to do.
Let $\phi:R^*\to R$ be a group isomorphism. Let's prove that $1+1=0$. In fact, if $1+1\ne0$, then $0=\phi(1)=\phi((-1)^2)=(1+1)\phi(-1)$, and therefore $\phi(-1)=0=\phi(1)$. Therefore, $-1=1$ as well. Therefore $1+1=0$.
Then, for all $x\in R^*$, we have $\phi(x^2)=\phi(x)+\phi(x)=0$. Therefore $x^2=1$ for all $x\in R^*$. But this means $(x-1)(x+1)=0$, i.e. $(x-1)^2=0$. Therefore, $x=1$.
This means that $R^*=\{1\}$. Since the zero ring isn't a domain, $\lvert R\rvert\ge2>\lvert R^*\rvert$.