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I'm attempting some questions from Abstract Algebra Theory and Applications by Judson (found here) and this one is a bit problematic.

Let $S = R\setminus \{−1\}$ and define a binary operation on $S$ by $a \ast b = a + b + ab$. Prove that $(S,\ast)$ is an abelian group.

See, ordinarily, I'd look at $\ast$ and just switch the roles of a and b and see that it is abelian but the negative reals are a potential problem here especially because they only left out $-1$.

I don't see why $-1$ is an obstacle to this being a binary operation or an abelian one and even if it were, I am not seeing how I can offer a proof that is satisfactory for $S$ as defined. Any hints are appreciated.

Shaun
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EddieMath
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  • Look up “transport of structure” in this site – lhf Dec 02 '21 at 22:50
  • If you're worried about the closure, you can show very simply that the only way to have $a * b = -1$ is to have $a = -1$ or $b = -1,$ so for any pair of $a$ and $b$ in $S$ we would have $a * b \in S.$ – Stephen Donovan Dec 02 '21 at 22:58
  • https://math.stackexchange.com/questions/3773328/show-bbb-r-setminus-%E2%88%921-is-an-abelian-group-where-ab-abab-solve This question was asked here. – George Dec 02 '21 at 23:08
  • See https://math.stackexchange.com/questions/1971224/let-g-mathbb-r-together-with-the-operation-xy-x-y-xy-forall-x-y/1971269#1971269 – lhf Dec 02 '21 at 23:46

2 Answers2

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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 $(\mathbb{R} \setminus\{0\}, \times)$.

Sera Gunn
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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 $\ast$.

Shaun
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