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For $S_6$,the possible orders are 1,2,3,4,5,6.

While for $A_6$, the possible orders are 1,2,3,4,5 but not 6.

Why does $A_6$ neglects the 6-cycle in $S_n$ which also has 0 number of 2-cycles and hence an even permutation .

I was going to upload the image of possible orders of $S_6$. but error pops up that I can't upload it .

  • You are confusing the order of a permutation with the sign of the permutation, which is given by: those that can only be expressed with an odd number of transpositions, and those that can only be expressed with an odd number of transpositions. $(1, 2, 3, 4, 5, 6) \in S_6$ is odd. $(1,2,3,4,5) \in S_6$ is even, hence in $A_6$. Please review your notes on transpositions (or permutations written in 2-cycles that are not necessarily disjoint). – amWhy Nov 06 '17 at 19:04
  • @amWhy pardon me sir if I am missing something again ,but I want to ask that $1,2,3,4,5,6$ is a 6-cycle and has no 2-cycle in it . $A_6$ will contain every possible arrangement such that number of 2-cycles is even. 6-cycle has even no of 2-cycle thus it must be in $A_6$. I am distributing possible arrangement in $S_6$as 6-cycle,5-cycle and one fixed point, 4-cycle and 2fixed points ,4-cycle and one transposition (2-cycle) etc. $1,2,3,4,5$has form of a 5-cycle and fixed point thus even no. of 2-cycle and belongs to $A_6 $. –  Nov 06 '17 at 19:16
  • In terms of a single cycle, order of the permutation is its length. But the number of numbers in the single cycle does not determine even or odd-ness. Every permutation (including single-cycles), can be expressed in terms of an even or an odd number of transpositions. For example, in the question I link to, there are at least two algorithms (or ways) in which to express, e.g. (1, 2, 3, 4, 5, 6) as a product of (usually not disjoint) transpositions: (12)(23)(34)(45)(56). Count the transpositions: there are five, hence the six cycle we started with is odd, and hence, not an element in $A_6$. – amWhy Nov 06 '17 at 19:24
  • The same odd number of transpositions results if we use a similar, but different way of decomposing the 6-cycle into transpositions: $(123456) = (16)(15)(14)(13)(12)$. Again, we have an odd number of transpositions. An transposition that is decomposed into an odd number of transpositions, can never be decomposed in any way, to an even number of transpositions. – amWhy Nov 06 '17 at 19:27
  • @amWhy Thanks sir ,I just remembered that only a cycle(single)with odd length can generate Even Number of 2-cycle .(1,2,3,4,5,6) is of order even and hence cannot . Thanks sir . –  Nov 06 '17 at 19:28
  • Glad to have helped. – amWhy Nov 06 '17 at 19:30

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