Prove groups $(\mathbb{R}_+,\times)$ and $(\mathbb{R}\setminus\{0\},\times)$ are not isomorphic [closed]

Question

Prove that group of positive real numbers with multiplication is not isomorphic to group of real numbers without zero with multiplication

Answer 1

Suppose there existed an isomorphism $$\varphi : \mathbb{R}^* \rightarrow \mathbb{R}_+$$ it would be in particular a bijection. However it should also satisfy that $\varphi (ab)=\varphi (a) \varphi(b)$. So let's examine $\varphi(-1)$.

$$\varphi (1)= \varphi((-1)(-1)) = (\varphi(-1))^2.$$ Now, since this is a homomorphism, it should be true that $$\varphi (1) = 1,$$ so $$(\varphi(-1))^2 = 1$$ since $\varphi(-1) \in \mathbb{R}_+$, $$\varphi(-1)=1,$$ so it would not be an injection, hence it would not be an isomorpshim.

Answer 2

Any homomorphism $h:\mathbb{R}_+ \longrightarrow \mathbb{R}\setminus\{0\}$ clearly satisfies $h(1)=1$. Moreover, we claim $h(b)>0$ for all $b\in \mathbb{R}_+$.

Indeed, for all $a\in \mathbb{R}_+$, $h(a^2)={h(a)}^2>0$. It suffice to note that every $b\in\mathbb{R}_+$ there is some $a\in\mathbb{R}_+$ with $a^2=b$.

It follows that $h$ can never be surjective, and hence cannot be an isomorphism.

Answer 3

How many solutions are there to the equation $x^2=1$ in both groups?

Note that $G$ and $G'$ are isomorphic, they have to have the same number of solutions to $x^2=1$. Let $\varphi: G \to G'$ be an isomorphism. Suppose that $a^2=1$ in $G$. Then $1=\varphi(1)=\varphi(a^2)=\varphi(a)^2$, so $\varphi(a)$ is a solution to $x^2=1$ in $G'$. Conversely, if $a^2=1$ in $G'$, consider the inverse isomorphism $\varphi^{-1}: G' \to G$. We have $1=\varphi^{-1}(1)=\varphi^{-1}(a^2)=\varphi^{-1}(a)^2$, so $\varphi^{-1}(a)$ is a solution to $x^2=1$ in $G$. So for every solution to $x^2=1$ in one of the groups, there is a (unique - because $\varphi, \varphi^{-1}$ are isomorphisms) corresponding solution to $x^2=1$ in the other group. Thus we obtain a 1-1 correspondence between solutions to $x^2=1$ in $G$ and solutions to $x^2=1$ in $G'$.