Isomorphism between groups of real numbers Is the group $\left(\Bbb{R}^*, \cdot\right)$ isomorphic to the group $\left(\Bbb{R},+\right)$ ?
I think that they are not but not sure how to show it. 
How would I define $\phi$?
 A: $x^2=y$ does not have a solution in $\mathbb{R}^\star$ when $y=-1$, yet $2x=y$ may be solved for every element $y\in\mathbb{R}$.
A: Hint 1: Suppose $\phi : \left(\Bbb{R},+\right)\to\left(\Bbb{R}^{\times},\cdot\right)$ is such an isomorphism, and consider the equation $x/2 + x/2 = x$ in $\left(\Bbb{R},+\right)$ (which holds for any $x\in\Bbb{R}$). What does this imply about the equation under the image of $\phi$ (can it still hold when $\phi(x) < 0$)?
Hint 2: Perhaps an isomorphism can be constructed between $\left(\Bbb{R}^+,\cdot\right)$ and $\left(\Bbb{R},+\right)$. Consider the map
\begin{align*}
\phi:\Bbb{R}&\to\Bbb{R}^+\\
x&\mapsto e^x.
\end{align*}
Can you use the properties of exponentials to show this is a bijective homomorphism of groups? (Also, note that if you show $\left(\Bbb{R}^+,\cdot\right)\cong\left(\Bbb{R},+\right)$, then you have shown $\left(\Bbb{R},+\right)\cong \left(\Bbb{R}^+,\cdot\right)$).
A: hint 3: $(\mathbb{R}^*,\cdot)$ has a nontrivial normal subgroup $\{1,-1\}$ (the kernel of the absolute value endomorphism $\operatorname{abs}$). any nontrivial subgroup of $(\mathbb{R},+)$ contains a copy of $(\mathbb{Z},+)$
formally, assume $\phi:(\mathbb{R},+)\rightarrow(\mathbb{R}^*,\cdot)$ is an isomorphism, and set $x=\phi^{-1}(-1)$. then
$$
(\operatorname{abs}\circ\phi)(x+x)=1\cdot1=1
$$
hence
$$
x+x\in\{0,x\}=\phi^{-1}\{1,-1\}
$$
a contradiction
