By Thomas Markwig Keilen

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**Additional resources for Algebraic Structures [Lecture notes]**

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Thus sgn(σ) = − sgn(σ ′ ) = sgn(σ ′ ) · sgn(τ). 14 every permutation is a product of transpositions of consecutive integers. 40 Let σ1 = τ˜1 ◦ · · · ◦ τ˜r and σ2 = τ˜r+1 ◦ · · · ◦ τ˜r+s be given as products of such transpositions of consecutive numbers. By induction on r + s we see that sgn(σ1 ◦ σ2) = (−1)r+s = (−1)r · (−1)s = sgn(σ1) · sgn(σ2). , and b. follows by induction on k. For c. let σ = τ1 ◦ · · · ◦ τk = τ1′ ◦ · · · ◦ τl′ with transpositions τi, τj′ ∈ Sn. Then by b. (−1)k = sgn(σ) = (−1)l, and thus either k and l are both even or both odd.

Proof: Let σ = σ ′ ◦ τ ∈ Sn with σ ′ ∈ Sn and τ = (i i + 1) for a i ∈ {1, . . , n − 1}. Is (i, i + 1) an error pair of σ ′ then τ cancels this one our and σ has one error pair less than σ ′ . Is conversely (i, i + 1) not an error pair of σ ′ then the composition with τ creates this error pair and σ has one error pair more than σ ′ . Thus sgn(σ) = − sgn(σ ′ ) = sgn(σ ′ ) · sgn(τ). 14 every permutation is a product of transpositions of consecutive integers. 40 Let σ1 = τ˜1 ◦ · · · ◦ τ˜r and σ2 = τ˜r+1 ◦ · · · ◦ τ˜r+s be given as products of such transpositions of consecutive numbers.

Thus m is equivalent to r, and therefore m ∈ r, where r is one of the above n equivalence classes. It remains to show for 0 ≤ i < j ≤ n − 1 that i = j. Suppose i = j then j would be equivalent to i and hence j − i ∈ nZ would be a multiple of n. By assumption, 18 Note here that the group operation is addition, so that a left coset is not denoted by “g · U” but by “g+U”. Maybe this would not be so deceiving if not at the same time the subgroup U = nZ itself looked like a multiplicative left coset — which it is not!