Show G is abelian....

• Mar 17th 2014, 04:49 AM
Aman3230
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.
• Mar 17th 2014, 07:44 AM
FernandoRevilla
Re: Show G is abelian....
Quote:

Originally Posted by Aman3230
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 $a,b\in G.$

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Quote:

Give an example to show that this result does not holds forsemi groups.