In a algebric Non abelian group ,do the product of 4 elements always give the same result, or it depends upon the order in which elements are taken... I think order is immaterial, What u say frnds...

The whole point of nonabelian groups is that the order does matter! Consider the simplest nonabelian group there is, the permutation group of \(3\) symbols. Then
\[ (12)(123)(12)(13) \; =\; (12) \qquad (12)(12)(123)(13) = (23) \]
(These permutation act "from the left", so \((13)\) is the permutation that acts first in the above products.)

Here's another example, which works in any group. Since \(G\) is nonabelian, find elements \(a , b\) with \(ab \neq ba\). If \(e\) is the identity element, then
\[ eabe \neq ebae \]

I think Sir i havent properly communicated my problem. I think in a non abelian group,. The product of 4 elements of the group is same irrespective of the order u choose...

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TopNewestThe whole point of nonabelian groups is that the order does matter! Consider the simplest nonabelian group there is, the permutation group of \(3\) symbols. Then \[ (12)(123)(12)(13) \; =\; (12) \qquad (12)(12)(123)(13) = (23) \] (These permutation act "from the left", so \((13)\) is the permutation that acts first in the above products.)

Here's another example, which works in any group. Since \(G\) is nonabelian, find elements \(a , b\) with \(ab \neq ba\). If \(e\) is the identity element, then \[ eabe \neq ebae \]

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Yes order in general not possess any relationship with output in non abelian group

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I think Sir i havent properly communicated my problem. I think in a non abelian group,. The product of 4 elements of the group is same irrespective of the order u choose...

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You mean the order in which you do the multiplication. That is true. For example \[ (ab)(cd) = (a(bc))d\] Multiplication is associative.

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One of the group axioms is associativity so it will hold for a product of any number of elements.

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