# Prove that the reals are isomorphic to the multiplicative group if reals with this group operation

If $G = \mathbb R \backslash \{-1\}$ prove $G$ is a group and that it is isomorphic to the multiplicative group of the non zero real numbers.
The group operation of $G$ is $a*b = a + b + ab$
I proved that $G$ is a group because its identity element is zero, every element has an inverse, mainly $\frac {-a}{1+a}$ however, to prove the isomorphism, I know that $\phi (0) = 1$ because identities map to identities, and $\phi (\frac {-a}{1+a}) = \frac 1a$
Yet, setting $a=1$ $\phi(0) = \phi (-1/2)=1$ proving this is not a bijection and therefore not an isomorphism. What am I doing wrong?

• Your map is contradictory: you say $\phi(0)=1$, but $\phi(0)=\phi(\frac{-0}{1+0})=\frac{1}{0}\neq 1$. – Joffysloffy Nov 23 '15 at 18:50
• So what map would be correct? Is that not the inverse? – Guacho Perez Nov 23 '15 at 18:51
• I do not know yet; I only noticed this inconsistency. – Joffysloffy Nov 23 '15 at 18:52

Take the map $\phi:a\mapsto a+1$. Then
\begin{aligned} \phi(a*b)&=\phi(a+b+ab)\\ &=a+b+ab+1\\ &=(a+1)(b+1)\\ &=\phi(a)\phi(b). \end{aligned} So $\phi$ is a homomorphism.
It's injective, because if $a\in\ker\phi$, then $\phi(a)=1$, hence $a+1=1$, which means that $a=0$.
It is surjective: let $b\in\mathbb{R}\setminus\{0\}$, then $b-1\in G$ and $\phi(b-1)=b-1+1=b$.
I basically took your idea of assuring inverses are sent to inverses and simply tried $\phi(\frac{-a}{1+a})=\frac{1}{a+1}$ to avoid division by zero. This resulted in the map $a\mapsto a+1$.
• Nice! But isn't the inverse of $a$ in the multiplicative group of non zero reals equal to $1/a$? – Guacho Perez Nov 23 '15 at 19:15
• Thank you :). It is, but you don't have to send the inverse of one element in $G$ to the element denoted by the same value in $\mathbb{R}\setminus\{0\}$. ‘Sending inverses to inverses’ means that if $a$ is sent to $b$, then $a^{-1}$ is sent to $b^{-1}$, and that is what happens. – Joffysloffy Nov 23 '15 at 19:18
• You're welcome :)! Also, just realized that you were actually just using the identity (take the inverse on both sides, you get $\phi(a)=a$). – Joffysloffy Nov 23 '15 at 19:24