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Math Help - Show G is abelian....

  1. #1
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    Exclamation Show G is abelian....

    G is a group in which (ab)i = aibifor three consecutive integers i
    and for every a,b Є G. Show that G is abelian.

    Give an example to show that this result does not holds forsemi groups.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Show G is abelian....

    Quote Originally Posted by Aman3230 View Post
    G is a group in which (ab)i = aibifor three consecutive integers i
    and for every a,b Є G. Show that G is abelian.
    Suppose

    $$(1)\qquad (ab)^{k-1}=a^{k-1}b^{k-1}$$ $$(2)\qquad (ab)^{k}=a^{k}b^{k},$$ $$(3)\qquad (ab)^{k+1}=a^{k+1}b^{k+1}.$$

    Using $(1)$ and $(2):$ $$\displaystyle\begin{aligned}
    &(ab)^{k-1}=a^{k-1}b^{k-1}=a^{-1}a^kb^kb^{-1}=a^{-1}(ab)^kb^{-1}\\
    &=a^{-1}(ab)(ab)\ldots (ab)b^{-1}=(ba)(ba)\ldots (ba)=(ba)^{k-1}.
    \end{aligned}$$ Using $(2)$ and $(3):$ $(ab)^{k}=a^{k}b^{k}$ y $(ab)^{k+1}=a^{k+1}b^{k+1}:$
    $$\displaystyle\begin{aligned}
    &(ab)^{k}=a^{k}b^{k}=a^{-1}a^{k+1}b^{k+1}b^{-1}=a^{-1}(ab)^{k+1}b^{-1}\\
    &=a^{-1}(ab)(ab)\ldots (ab)b^{-1}=(ba)(ba)\ldots (ba)=(ba)^{k}.
    \end{aligned}$$ Using $(ab)^{k-1}=(ba)^{k-1}:$ $$\left \{ \begin{matrix} (ab)^k=(ab)^{k-1}(ab)\\ (ba)^k=(ba)^{k-1}(ba)=(ab)^{k-1}(ba)\end{matrix}\right.$$ and by the equality $(ab)^{k}=(ba)^{k},$ we have $(ab)^{k-1}(ab)=(ab)^{k-1}(ba).$

    By the cancellation law: $ab=ba$ for all $a,b\in G.$

    
    Give an example to show that this result does not holds forsemi groups.
    Please, show some work.
    Thanks from Deveno and johng
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