Maybe its my flu...but what do you mean exactly? What I said are two definitions out of the book.HallsofIvy said:So, it appears you are saying that all the permutations involved in [itex]a^{-1}b^{-1}ab[/itex] is a two-cycle and this is a product of 4 of them!
A permutation is considered even if it can be written as an even number of transpositions. A transposition is a permutation that swaps two elements in a set. In other words, an even permutation can be broken down into an even number of swaps or rearrangements.
To determine if a permutation is even or odd, you can count the number of inversions. An inversion occurs when two elements in a permutation are in reverse order. If the number of inversions is even, then the permutation is even. If the number of inversions is odd, then the permutation is odd.
The proof for (a^-1)(b^-1)(a)(b) being an even permutation involves showing that it can be broken down into an even number of transpositions. This can be done by considering the individual elements of the permutation and analyzing how they are affected by the transpositions. By breaking down (a^-1)(b^-1)(a)(b) into transpositions, we can see that it can be written as an even number of swaps, thus making it an even permutation.
The proof for (a^-1)(b^-1)(a)(b) being an even permutation directly relates to the definition of an even permutation. By showing that (a^-1)(b^-1)(a)(b) can be written as an even number of transpositions, we are demonstrating that it can be broken down into an even number of swaps. This aligns with the definition of an even permutation, which is a permutation that can be written as an even number of transpositions.
Yes, this proof can be applied to any permutation in Sn. This is because the proof relies on the properties of permutations and their ability to be broken down into transpositions. Therefore, the proof can be applied to any permutation in Sn, regardless of the specific elements or order of the permutation.