# Thread: Proving that this is a group

1. ## Proving that this is a group

Hello, I would really appreciate any help with the following problem:

Prove that $(G, *)$ is a group, where $G=<-1,1>$ and $a*b=\dfrac{a+b}{1+ab}$,.... $\forall a, b \in G$.

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To prove that this is a group, $(G,*)$ must satisfy closure, associativity, existence of an identity element and existence of inverse elements.

Could the property of associativity be demonstrated like this?

$(a*b)*c=(\dfrac{a+b}{1+ab})*c=...=\dfrac{a+b+c+abc }{1+ab+ac+bc}$

$a*(b*c)=a*(\dfrac{b+c}{1+bc})=...=\dfrac{a+b+c+abc }{1+ab+ac+bc}$

Therefore, $(a*b)*c=a*(b*c)$, so the property of associativity is satisfied.

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Is the identity element $0$?

From $a*0= \dfrac{a+0}{1+a \cdot 0}=0*a=a$

it follows $a*0=0*a=a$

_______________________________

Is the inverse element of $a$, $-a$?

Because $a*(-a)=\dfrac{a-a}{1-a^2}=\dfrac{0}{1-a^2}=(a \in <-1,1>)=(-a)*a = 0$

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If the above is correct (and I'm not certain that it is), the only property left to be demonstrated is closure, i.e. if $a, b \in <-1,1>$ then $a*b =\dfrac{a+b}{1+ab} \in <-1,1>$.

And here I need your help! How to demonstrate the property of closure?

There is a hint in the text: "Observe that $|1+ab|=1+ab$, and prove that $|a+b| \leq 1+ab$ if and only if $0 \leq (1-a)(1-b)$ and $0 \leq (1+a)(1+b)$".

So, how to prove the statements from the hint, and, more importantly, how to apply them to demonstrate the closure property of $(G,*)$?

Many thanks!

2. Originally Posted by gusztav
Hello, I would really appreciate any help with the following problem:

Prove that $(G, *)$ is a group, where $G=<-1,1>$ and $a*b=\dfrac{a+b}{1+ab}$,.... $\forall a, b \in G$.

_______________________________

To prove that this is a group, $(G,*)$ must satisfy closure, associativity, existence of an identity element and existence of inverse elements.

Could the property of associativity be demonstrated like this?

$(a*b)*c=(\dfrac{a+b}{1+ab})*c=...=\dfrac{a+b+c+abc }{1+ab+ac+bc}$

$a*(b*c)=a*(\dfrac{b+c}{1+bc})=...=\dfrac{a+b+c+abc }{1+ab+ac+bc}$

Therefore, $(a*b)*c=a*(b*c)$, so the property of associativity is satisfied.

_______________________________

Is the identity element $0$?

From $a*0= \dfrac{a+0}{1+a \cdot 0}=0*a=a$

it follows $a*0=0*a=a$

_______________________________

Is the inverse element of $a$, $-a$?

Because $a*(-a)=\dfrac{a-a}{1-a^2}=\dfrac{0}{1-a^2}=(a \in <-1,1>)=(-a)*a = 0$

_______________________________

If the above is correct (and I'm not certain that it is), the only property left to be demonstrated is closure, i.e. if $a, b \in <-1,1>$ then $a*b =\dfrac{a+b}{1+ab} \in <-1,1>$.

And here I need your help! How to demonstrate the property of closure?

There is a hint in the text: "Observe that $|1+ab|=1+ab$, and prove that $|a+b| \leq 1+ab$ if and only if $0 \leq (1-a)(1-b)$ and $0 \leq (1+a)(1+b)$".

So, how to prove the statements from the hint, and, more importantly, how to apply them to demonstrate the closure property of $(G,*)$?

Many thanks!

First ,what do you mean by $G=<-1,1>$ ?? The subset of the reals (or complex) with those two elements? If so zero cannot be anything there since it doesn't belong to the set!
Also, the product of $1\,,\,-1$ isn't defined: $1*-1:= \frac{1+(-1)}{1+1(-1)}=\frac{0}{0}$ ...!!

If this is so the above isn't a group.

But perhaps you meant the open interval $(-1,1)$ which, by some misterious reason, you denote by $<-1,1>$...(!)

Then you must show $a,b\in(-1,1)\Longrightarrow \frac{a+b}{1+ab}\in (-1,1)\Longleftrightarrow \left|\frac{a+b}{1+ab}\right|<1$ $\Longleftrightarrow (a+b)^2<(1+ab)^2\Longleftrightarrow a^2+2ab+b^2

$\Longleftrightarrow a^2(b^2-1)-(b^2-1)>0\Longleftrightarrow (a^2-1)(b^2-1)>0$ , and this last inequality is obviously true.

Tonio

3. Thank you very much, Tonio!

Originally Posted by tonio
But perhaps you meant the open interval $(-1,1)$
Yes, I meant the open interval $\{x \in \mathbb{R} : -1

Originally Posted by tonio
Then you must show $a,b\in(-1,1)\Longrightarrow \frac{a+b}{1+ab}\in (-1,1)\Longleftrightarrow \left|\frac{a+b}{1+ab}\right|<1$ $\Longleftrightarrow (a+b)^2<(1+ab)^2\Longleftrightarrow a^2+2ab+b^2

$\Longleftrightarrow a^2(b^2-1)-(b^2-1)>0\Longleftrightarrow (a^2-1)(b^2-1)>0$ , and this last inequality is obviously true.
Aha! Now everything is clear.
(Because $a \in (-1,1) \Longrightarrow |a|<1 \Longrightarrow a^2<1 \Longrightarrow a^2-1<0$ and, similarly, $b^2-1<0$, so $(a^2-1)( b^2-1)>0$ )