Show that the Group of Real Numbers Excluding $-1$ is Isomorphic to $\mathbb{R}^*$

Question

Let $\langle S, \cdot\rangle$ be the group of all real numbers excluding $-1$ where $a \cdot b = a + b + ab$. Show that $\langle S, \cdot\rangle$ (which will now be shortened to $S$) is isomorphic to $\mathbb{R}^*$, the group of nonzero real numbers.

Proof:

Let $\phi: \mathbb{R}^* \to S$ and $\phi (a) = ax$ where $x \in S, \mathbb{R}^*$

Thus, $\phi$ is injective because $\phi(a) = \phi(b) \implies a\cdot x = b\cdot x \implies a=b$ (by right cancellation)

Now, let $y \in S$. Thus, $\phi(y x^{-1}) = y\cdot x^{-1} \cdot x = y$. So $\phi$ is surjective

Finally, $\phi(ab) = a\cdot b\cdot x$...

That final step is where I got stuck. I'm struggling to show that $\phi$ is a homomorphism. I've tried a couple of different functions, but I just can't seem to find a homomorphism between $S$ and $\mathbb{R}^*$.

One idea that I came up with to find a homomorphism was by assuming the homomorphism exists and manipulating the equation:

$$\phi(ab) = \phi(a) \cdot \phi(b)$$ $$\phi(ab) = \phi(a) + \phi(b) + \phi(a) \phi(b)$$

but this didn't help me much (unless I'm failing to realize something).

Perhaps I could start with $\phi(a) \cdot \phi(b) = (a + x + ax) \cdot (b + x + bx)$, but I'm not sure how I would go about computing the right hand side of the equation any further. How can I finish the proof (or find a different bijection)?

Answer 1

The isomorphism from $S$ to $\mathbb{R}^*$ is $x\mapsto x+1$. Prove that this is an isomorphism.

Answer 2

Note that $(a+1)(b+1)=a+b+ab+1$. In fact, if you define $f(k;a,b)=a+b+kab$ (where $k$ is a real number), you can then show that there always exists an isomorphism to the real number under multiplication. An intuition/application of this in probability is to take the union of two probabilities.