Can we find a field, if we have a group?

Question

1. Lets be $(G,\oplus)$ a group isomorphic to $(\mathbb{R},+)$. Can we find an operator "$\odot$" so that $(G,\odot,\oplus)$ is field?
2. Lets be $(G,\odot)$ a group isomorphic to $(\mathbb{R}^\ast,\cdot)$. Can we find an operator "$\oplus$" and a number "$e$" so that $(G\cup\{e\},\odot,\oplus)$ is field?
3. Is there any bibliography about these properties?

For example: Lets be $a\odot b=5ab$ for all $a,b\in\mathbb{R}^\ast$. Then it is easy to see that we can find an operator "$\oplus$" and a number "$e$" ("$\oplus$" is "$+$" and $e=0$) so that $(\mathbb{R}^\ast \cup \{e\},\odot,\oplus)$ is field. But what about other choices? If "$\odot$" was "$+$" or $a\odot b=a+b+ab$, then what?

Answer 1

1. Let $f:G\rightarrow\Bbb R$ be an isomorphism of additive groups. Define $$\odot:G\times G\xrightarrow{f\times f}\Bbb R\times\Bbb R\xrightarrow{\cdot}\Bbb R\xrightarrow{f^{-1}}G.$$ We can check that this makes $G$ a field, with multiplicative unit $f^{-1}(1)$.

2. Let $f:G\rightarrow\Bbb R^*$ be an isomorphism of multiplicative groups. Let $e$ be a symbol, and extend $\odot$ to $G\cup\{e\}$ by $g\cdot e=e\cdot g=e$. Then $G\cup\{e\}$ is a monoid. Extend $f$ to $(G\cup\{e\})$ by $f(e)=0$. $f$ is now an isomorphism of monoids, from $G\cup\{e\}$ to $\Bbb R^*\cup\{0\}=\Bbb R$. Define $$\oplus:(G\cup\{e\})\times (G\cup\{e\})\xrightarrow{f\times f}\Bbb R\times\Bbb R\xrightarrow{+}\Bbb R\xrightarrow{f^{-1}}(G\cup\{e\}).$$ We can check that this makes $(G\cup\{e\})$ a field, with additive unit $e$.

Answer 2

What if you set $g_1 \odot g_2 = f^{-1}[f(g_1)\cdot_{\mathbb{R}}f(g_2)]$

and

$g_1 +_G g_2 = f^{-1}[f(g_1)+_{\mathbb{R}} f(g_2)] $

where $f$ is the isomorphism.