1
$\begingroup$

Let $G = (1, 2)$ and $x * y = \frac{3xy - 4x - 4y + 6}{2xy - 3x - 3y + 5}$, $\forall \, x, y \in G$. Check if $(\mathbb{R}_+^*, \cdot)$ and $(G, \cdot)$ are isomorphic through a function $f : (0, \infty) \to G$, $f(x) = \frac{x + \alpha}{x + \beta}$, where $\alpha$ and $\beta$ are real numbers.

We have two conditions for isomorphisms: $f$ must be bijective and $f(xy) = f(x) * f(y)$, $\forall \, x, y \in \mathbb{R}_+^*$. Using the second condition is definitely an overkill and very error-prone. Is there another way to find isomorphisms?

Thank you!

  • 0
    Have you found the identity element of $G$? – 2017-02-10
  • 0
    Yes, it's $3/2$. – 2017-02-10
  • 2
    Just an idea: can you get anything out of the fact that $f(1)=\frac{1+\alpha}{1+\beta}=3/2$ ? – 2017-02-10

1 Answers 1

1

Notice that such an $f$ has to be continuous, and it must be injective, so it must be monotonous; since $\lim_{x\to \infty} f(x)=1$, you must have $\lim_{x\to 0}f(x)=2$, and thus $\alpha=2\beta$. With the added condition suggested by Sloan in the comments that $f(1)=\frac{3}{2}$, we get that $2+4\beta=3+3\beta$, hence $\beta=1$ and $\alpha=2$.

Now you need to check that this $f$ is really a homomorphism, which will probably be quite tedious, given the formula for the product in $G$. Instead you can first compute that the inverse of $f$ is given by $g(x)=\frac{2-x}{x-1}$; and $f$ is a homomorphism if and only if $g$ is one. This condition is in turn equivalent to check that $$x\ast y=f(g(x)\cdot g(y)).$$Now the right-hand side in this equation is given by $$f\left(\frac{2-x}{x-1}\cdot\frac{2-y}{y-1}\right)=f\left(\frac{4-2x-2y+xy}{xy-x-y+1}\right)=\dfrac{\frac{4-2x-2y+xy}{xy-x-y+1}+2}{\frac{4-2x-2y+xy}{xy-x-y+1}+1}=\frac{3xy-4x-4y+6}{2xy-3x-3y+5},$$which is indeed equal to $x\ast y$.