Show that $G$ is a group under $*$

Question

Let $G$ be the set of rational numbers $x$ with $x \neq\frac{-1}{2}.$ For $x, y ∈ G$ define $$x ∗ y = 2xy + x + y.$$

Show that $G$ is a group under $*$.

I know how to show that associativity holds and that the identity $e$ is zero. However, I'm struggling to show that the group is closed under taking inverses.

I know I need to show that there exists an element $x^{-1}$ so that $x*x^{-1}=0.$ So this leads to $$2xx^{-1}+x+x^{-1}=0$$

What do I do next?

Answer 1

Consider the map $\phi: G \to \mathbb Q^*$ given by $\phi(x)=1+2x$.

The operation $*$ in $G$ is just the pullback of ordinary multiplication in $\mathbb Q^*$: $$ x * y = \phi^{-1}(\phi(x)\phi(y)) $$

Therefore, $(G,*)$ is a group, isomorphic to $\mathbb Q^*$.

In particular, the inverse of $x \in G$ with respect to $*$ is $\phi^{-1}(\phi(x)^{-1}))=\phi^{-1}\left(\dfrac{1}{1+2x}\right)=\dfrac{-x}{1+2x}$.

Answer 2

You can solve this to see that

$$x^{-1}=\frac{-x}{2x+1}$$

which is valid because $x\neq\frac{-1}{2}$.