Abstract Algebra

**mathman88** · March 9th 2009, 08:19 AM

Show that $(\mathbb{R}, \; +)$ is not isomorphic to $(\mathbb{R}^*, \; \cdot)$ , where $\mathbb{R}^*$ is the group of units in $\mathbb{R}$ . i.e. $\mathbb{R}^* = \mathbb{R}\backslash \{0\}$

**red_dog** · March 9th 2009, 08:37 AM

Suppose that $(\mathbf{R},+)$ is isomorphic to $(\mathbf{R}^*,\cdot)$

Let $f<img src=$ \mathbf{R},+)\to (\mathbf{R}^*.\cdot)" alt="f

\mathbf{R},+)\to (\mathbf{R}^*.\cdot)" /> an isomorphism.

Then $f(x+y)=f(x)f(y), \ \forall x,y\in\mathbf{R}$

For $y=x\Rightarrow f(2x)=(f(x))^2\Rightarrow f(x)=\left(f\left(\frac{x}{2}\right)\right)^2\Righ tarrow f(x)\geq 0, \ \forall x\in\mathbf{R}$

Then f is not a surjection.

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