Finding an isomorphism from $\mathbb{R}^\times$ to a defined group $G$

Question

Here's the problem I am solving:

$G=\{x\in \mathbb{R}:x\not = 0\}$. The operation for $G$ is "$*$", with $x*y=\frac{1}{2}xy.\mathbb{R}^\times$ is the multiplicative group $\mathbb{R}.$ Find an isomorphism from $\mathbb{R}^\times\rightarrow G.$

Clearly, $\mathbb{R}^\times$ is an abelian group. Upon examination, I concluded that $(G,*)$ is also an abelian group, with the identity $e=2$ satisfying $a*e=a\;\;\forall a\in G$.

I found that the equation $f(ab)=f(a)*f(b)=\frac{1}{2}f(a)f(b)$ is satisfied by $f(x)=2x\;\forall x\in \mathbb{R}^\times$.

The problem with $f(x)=2x$ is that it is not a bijection. It is surjective but not injective, as shown by the counterexample $0\in\mathbb{R}^\times,$ but $f(0)\not\in G.$

I cannot find a bijective $f$ satisfying this problem. My conclusion is that these groups are not isomorphic. But the textbook seems to imply that an isomorphism does exist, which is why I posed this question. Is my conclusion correct? If not, can you give me a hint on how to find $f$?

Thank you.

EDIT: I was unaware that the multiplicative group is defined as $\mathbb{R}^\times=\mathbb{R}\setminus\{0\},$ so I will close the question.

EDIT 2: Not sure how to close this question.

This is a special case of Showing that $G$ is a group under an alternative operation. — lhf, Sep 29 '14 at 15:47
$0\notin\mathbb R^*=\mathbb R\setminus{0}$. If you include zero, it's not a multiplicative group. Or are you using some other definition? — Joonas Ilmavirta, Sep 29 '14 at 15:48
@JoonasIlmavirta I didn't realize $\mathbb{R}^\times=(\mathbb{R}\setminus{0},\times)$. That answers the question. Also, I said that $f(x)=2x$ was not a bijection when in fact it is simply not a function at all. Thanks. — Patrick Shambayati, Sep 29 '14 at 15:52
$\Bbb R^\times$ isn't "the multiplicative group $\Bbb R$," it is the "multiplicative group of units of $\Bbb R$." In general, $R^\times$ is the collection of invertible elements of a ring $R$, which make a group under multiplication. — anon, Sep 29 '14 at 15:54
@whacka I see. I haven't studied rings yet, but I will keep that in mind — Patrick Shambayati, Sep 29 '14 at 15:57

Finding an isomorphism from $\mathbb{R}^\times$ to a defined group $G$

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