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I am trying to prove that $(G, *)$ is an abelian group with $G=(-1,1)$ and $a*b=$$\frac{x+y}{1+xy}$.

Thus far I have found that the identity element $e=0$. From here, I set $a*b=0$ and found $a^{-1}$ to be $-a$.

My work for trying to prove closure and that the set is abelian is:

Let $a,b \in G$, where $a=e$ and $b=a^{-1}$. Does $a*b=b*a$?

Evaluating, I see that $a^{-1}=a^{-1}$ which then proves that G is closed and G is an abelian group.

Did I go about this proof correctly and efficiently?

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    $\begingroup$ you mean $a*b=\frac{a+b}{1+ab}$? $\endgroup$ – Math137 May 7 '14 at 19:52
  • $\begingroup$ @Hagen von Eitzen Thank you. I understand. I delete it. $\endgroup$ – nadia-liza May 7 '14 at 20:11
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You have found neutral and inverse fine.

You need to show closure, i.e. if $-1<x,y<1$, is then also $-1<\frac{x+y}{1+xy}<1$? (To really be precise: Is the expression defined for all cases in the first place?)

Abelian is quite clear, because $x+y=y+x$ and $xy=yx$.

However, the toughest part (at least if one is forced to do direct computatons) is in fact associativity: This boils down to writing down the complicated fractions for $x*(y*z)$ and $(x*y)*z$ and showing that they are indeed equal.

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  • $\begingroup$ They aren't complicated at all. $\endgroup$ – egreg May 7 '14 at 20:17
  • $\begingroup$ @egreg Indeed, with the symmetic expression in your answer, associativity is clear. But at least this is about the only point that requires to actually write down and not just look at the expression ... :) $\endgroup$ – Hagen von Eitzen May 8 '14 at 5:38
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If $|a|<1$ and $|b|<1$, then also $|ab|<1$, so $-1<ab<1$ and $0<1+ab<2$. Consequently, $(a+b)/(1+ab)$ is defined for all $a,b\in (-1,1)$. Now, for $a,b\in(-1,1)$, $$ \frac{a+b}{1+ab}<1 $$ becomes $a+b<1+ab$ or $1-a-b+ab>0$, that is, $(1-a)(1-b)>0$, which is true. Similarly $$ \frac{a+b}{1+ab}>-1 $$ becomes $a+b>-1-ab$ or $(1+a)(1+b)>0$. Again this is true, under our hypotheses.

Let's tackle $a*(b*c)$; since $b*c=(b+c)/(1+bc)$, we have $$ a*(b*c)=\frac{a+\dfrac{b+c}{1+bc}}{1+a\dfrac{b+c}{1+bc}}= \frac{abc+a+b+c}{1+bc+ab+ac} $$ Can you try with $(a*b)*c$?

You computed the identity and the inverse correctly. However, being $G$ abelian means $$ a*b=b*a $$ for all $a,b\in G$.

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  • $\begingroup$ The last numerator should be $abc+a+b+c$, I believe. $\endgroup$ – Espen Nielsen May 7 '14 at 20:35
  • $\begingroup$ @espen180 Right. $\endgroup$ – egreg May 7 '14 at 20:36
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Another way to look at this is as follows. Note that $$tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ defines a bijective map from $\mathbb{R} \rightarrow (-1,1).$ And $$tanh(a+b)=\frac{tanh(a)+tanh(b)}{1+tanh(a)tanh(b)}.$$ And obviously addition carries over in the operation $*$ defined by the OP.

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The identity is indeed 0 and $a^{-1}$ is indeed $-a$. You have 3 things left to show:

Closure: You need to show that if a and b are in (-1, 1) then so is $\frac{a+b}{1+ab}$. It seems like the best way to do this would be to consider separate cases when a and b are positive or negative.

Associativity: You need to prove $a*(b*c) = (a*b)*c$. This should be straightforward, albeit messy algebra.

Commutativity: What you have done is show that the identity element commutes with group elements; you need to show an arbitrary group elements commutes with other group elements. This is clear, since $\forall a,b \in G$, $$a*b = \frac{a+b}{1 + ab} = \frac{b+a}{1 + ba} = b*a $$

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