No isomorphism between $(\mathbb R,+)$ and $(\mathbb R^*, \times)$ My goal is to disprove the existence of an isomorphism between $(\mathbb R,+)$ and $(\mathbb R^*, \times)$.
I proceeded by contradiction. Suppose $f$ is such a map.
Then $$f(0-0)=f(0)f(-0)=-f(0)^2$$
$$ f(0+0)=f(0)^2$$
$$f(0-0)=f(0+0)=f(0)$$
$$-f(0)^2=f(0)^2$$
thus, $$f(0)=0$$
This contradicts the fact that $f(0) \in \mathbb R^*$
Is this proof right ? It looks overly simple to me and doesn't use the one-to-one property of $f$.
EDIT: the proof is indeed flawed: check answers below.
For those looking for a valid one : let $x=f^{-1}(1)$ and $y=f^{-1}(-1)$. Then $x \neq y$ $$f(2x)=f(x)^2=1^2=(-1)^2=f(2y)$$
Since $f$ is one-to-one, $$2x=2y$$
$$x=y$$
 A: No. You used $f(-0)=-f(0)$ but this isn't true. The inverse in the multiplicative group is $x\mapsto \cfrac{1}{x}$. So $f(-0)=\cfrac{1}{f(0)}$

To find the "usual" proof, notice that $\exp$ is an isomorphism between $(\Bbb R,+)$ and $(\Bbb R_+^*, \times)$ but here, you have $(\Bbb R^*, \times)$ which is "bigger", approximately twice as "big". So maybe, you could find a property so that in the first group, you have only one element verifying that property and in the second group, you have two.
A: Other arguments may be:


*

*If $f : (\mathbb{R},+) \to (\mathbb{R}^*, \times)$ is a morphism, for all $x \in \mathbb{R}$ $$f(x)=f \left( 2 \frac{x}{2} \right)= f \left( \frac{x}{2} \right)^2 \geq 0,$$ hence $\mathrm{Im}(f) \subset \mathbb{R}_+^*$. In particular, $f$ cannot be an isomorphism.

*$(\mathbb{R},+)$ is a ($2$-)divisible group but not $(\mathbb{R}^*, \times)$ (for example, $-1$ has no square root).
A: I think the simplest possible argument (pretty much the same as the one you've ended up at) is as follows: the additive group is torsion-free, but in the multiplicative group, $(-1)^2=1$.
A slight modification of this argument works for any field (in fact, any domain): for any field $k$, if characteristic of $k$ is $0$, then $(k,+)$ is torsion-free, but $-1\neq 1$ and $(-1)^2=1$. If chracteristic of $k$ is $p>0$, then $1$ has order $p$ in $(k,+)$, but in the multiplicative group there is no element of order $p$: if $x^p=1$, then $0=x^p-1=(x-1)^p$ so $x=1$.
