# Permutations and Abelian Group proof.

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• Feb 17th 2010, 12:45 PM
Amer
$\displaystyle \beta=(2,3,9)(6,9,10)=\left(\begin{array}{cccccccc cc}1&2&3&4&5&6&7&8&9&10\\1&3&9&4&5&6&7&8&2&10\\1&3 &10&4&5&9&7&8&2&6\end{array}\right)$

$\displaystyle \beta=\left(\begin{array}{cccccccccc}1&2&3&4&5&6&7 &8&9&10\\1&3&10&4&5&9&7&8&2&6\end{array}\right)=(2 ,3,10,6,9)$

• Feb 17th 2010, 12:52 PM
MickQ
Quote:

Originally Posted by Amer
$\displaystyle \beta=(2,3,9)(6,9,10)=\left(\begin{array}{cccccccc cc}1&2&3&4&5&6&7&8&9&10\\1&3&9&4&5&6&7&8&2&10\\1&3 &10&4&5&9&7&8&2&6\end{array}\right)$

$\displaystyle \beta=\left(\begin{array}{cccccccccc}1&2&3&4&5&6&7 &8&9&10\\1&3&10&4&5&9&7&8&2&6\end{array}\right)=(2 ,3,10,6,9)$

Ah, I get it now! Thanks very much! (Happy)

So now, with Alpha*Eta = Beta,

that means Eta = Alpha^(-1) * Beta, how do you go about computing that?
• Feb 17th 2010, 01:19 PM
MickQ
After some calculations, would I be correct to state that:

Alpha^(-1) = Alpha

That's what I found. There what I need to compute is:

Eta = AlphaBeta
• Feb 18th 2010, 07:43 AM
Amer
Quote:

Originally Posted by MickQ
After some calculations, would I be correct to state that:

Alpha^(-1) = Alpha

That's what I found. There what I need to compute is:

Eta = AlphaBeta

ok it is easy now to solve it

$\displaystyle \eta = \alpha \beta$

$\displaystyle \eta = (1,2)(10,8)(2,3,10,6,9)$ simplify it
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