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Thread: Proving that this is a group

  1. #1
    Junior Member gusztav's Avatar
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    Proving that this is a group

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

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

    _______________________________

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

    Could the property of associativity be demonstrated like this?

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

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

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

    _______________________________

    Is the identity element $\displaystyle 0$?

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

    it follows $\displaystyle a*0=0*a=a$

    _______________________________

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

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

    _______________________________

    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 $\displaystyle a, b \in <-1,1>$ then $\displaystyle a*b =\dfrac{a+b}{1+ab} \in <-1,1>$.

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

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

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

    Many thanks!
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  2. #2
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    Quote Originally Posted by gusztav View Post
    Hello, I would really appreciate any help with the following problem:

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

    _______________________________

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

    Could the property of associativity be demonstrated like this?

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

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

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

    _______________________________

    Is the identity element $\displaystyle 0$?

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

    it follows $\displaystyle a*0=0*a=a$

    _______________________________

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

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

    _______________________________

    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 $\displaystyle a, b \in <-1,1>$ then $\displaystyle a*b =\dfrac{a+b}{1+ab} \in <-1,1>$.

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

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

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

    Many thanks!

    First ,what do you mean by $\displaystyle G=<-1,1>$ ?? 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 $\displaystyle 1\,,\,-1$ isn't defined: $\displaystyle 1*-1:= \frac{1+(-1)}{1+1(-1)}=\frac{0}{0}$ ...!!

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

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

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

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

    Tonio
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  3. #3
    Junior Member gusztav's Avatar
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    Thank you very much, Tonio!

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


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

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

    Thanks again for your help!
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