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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 $\phi$?

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Almost: One has an element of order 2, the other doesn't. Restrict your attention to positive reals under multiplication, and then you're right. How to show it? Exponentiation turns addition into multiplication; logarithms turn multiplication into addition. – anon Mar 28 '13 at 23:44
So am I correct in saying that it is isomorphic using the logarithm? – Charlie Brown Mar 29 '13 at 0:13
The groups you mention above are not isomorphic using the logarithm: the logarithm isn't a function from $\Bbb{R}\to\Bbb{R}$ after all, its domain is $\Bbb{R}^+$. However, look at the answers below to see a) how to show that the groups you mentioned above can't be isomorphic and b) how to find an isomorphism of groups that does work. – Stahl Mar 29 '13 at 0:18
$(\mathbb{R}^*,\cdot)$ is isomorphic to $\mathbb{R} \times (\mathbb{Z}/2\mathbb{Z})$ (as additive group), so it is not isomorphic to $(\mathbb{R},+)$. – Hanul Jeon Mar 29 '13 at 8:22

$x^2=y$ does not have a solution in $\mathbb{R}^\star$ when $y=-1$, yet $2x=y$ may be solved for every element $y\in\mathbb{R}$.

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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 $\left(\Bbb{R},+\right)\cong \left(\Bbb{R}^+,\cdot\right)$).

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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

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In other words, the multiplicative group has torsion, but the additive group is torsionfree (i.e. the characteristic is zero). – Martin Brandenburg Mar 29 '13 at 0:21

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