Group theory exercise from Judson text

Solution 1:

Let $f(x) = x + 1$ and $g(x) = x - 1$. Then what we have is $$x * y = (x + 1)(y + 1) - 1 = g(f(x)f(y))$$

Or, to write this in a way that might be easier to catch what is going on:

$$g(x)*g(y) = g(xy).$$

So $S$ is isomorphic to $(\mathbb{R} \setminus\{0\}, \times)$.

Solution 2:

Here $-1$ is an obstacle, essentially, because of the following.

Lemma: Let $e$ be the identity of a group $G$. Then the only idempotent of $G$ is $e$.

Proof: Let $x^2=x$. Then $ex=x=xx$. Multiplying on the right by $x^{-1}$ then gives $e=x$. $\square$

We have $-1\ast -1=-1-1+(-1)^2=-1$ and $0\ast 0=0+0+0^2=0$.

Moreover, we have that $a\ast a'=a+a'+aa'=0$ implies

$$a'=-\frac{a}{1+a},$$

where $a'$ is the inverse of $a$ with respect to $\ast$.