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If I know that the parity of a permutation is the parity of the number of transpositions, how can prove that the parity of the permutation is the parity of permutation decrement?

Permutation decrement is the difference between the number of truly movable elements and the number of independent cycles

Bernard
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Elias
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    What do you mean by "the parity of the permutation is the parity of permutation decrement"? Do you mean something like the parity of a $k$-cycle is the parity of $k-1$? – Clive Newstead Dec 23 '19 at 13:23
  • Maybe try: https://www.encyclopediaofmath.org/index.php/Permutation_of_a_set, https://math.stackexchange.com/questions/3074389/on-terminology-what-is-meaning-of-the-decrement-of-a-permuation-or-what-is – Alessio K Dec 23 '19 at 13:24

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I'm going to assume you're trying to prove that the parity of a $k$-cycle is equal to the parity of $k-1$. If this is the case, note that $$(a_1~~a_2~~\cdots~~a_{k-1}~~a_k) = (a_1~~a_k)(a_1~~a_{k-1}) \cdots(a_3~~a_1)(a_2~~a_1)$$ The expression on the right-hand side is a product of $k-1$ transpositions.

Clive Newstead
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