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?

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

  • 1
    $\begingroup$ Nice! But isn't the inverse of $a$ in the multiplicative group of non zero reals equal to $1/a$? $\endgroup$ – Guacho Perez Nov 23 '15 at 19:15
  • 1
    $\begingroup$ 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. $\endgroup$ – Joffysloffy Nov 23 '15 at 19:18
  • $\begingroup$ Oh, alright, thanks! Really good stuff! $\endgroup$ – Guacho Perez Nov 23 '15 at 19:19
  • 1
    $\begingroup$ You're welcome :)! Also, just realized that you were actually just using the identity (take the inverse on both sides, you get $\phi(a)=a$). $\endgroup$ – Joffysloffy Nov 23 '15 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.