Finding 12 Permutations Commuting with Alpha=(1,2,4,5)

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Homework Help Overview

The discussion revolves around finding 12 permutations in the symmetric group S6 that commute with the permutation alpha=(1,2,4,5). Participants are exploring the properties of permutations and the concept of commutativity in this context.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts various permutations and expresses uncertainty about their validity. Some participants point out that certain permutations are not distinct or do not commute with alpha. There is a discussion about the identity permutation and the implications of composing permutations.

Discussion Status

Participants are actively engaging with the problem, questioning the correctness of proposed permutations, and clarifying the definition of commutativity. Some guidance has been offered regarding the identity permutation and the use of fixed elements, but there is no explicit consensus on the total number of valid permutations.

Contextual Notes

There is mention of confusion regarding the definition of commuting permutations, and the original poster notes a discrepancy in the expected number of permutations. The original poster also indicates a search for additional resources to aid understanding.

SqrachMasda
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1) In S6 find 12 permutations that commute with alpha=(1,2,4,5)

i did (5,1,2,4)
(4,5,1,2)
(2,4,5,1)
?i don't know if transposition answers this?
but i also did
(1,5)(1,4)(1,2)
and similar for the ones above

i also used
(1,2,4,5)(2,1)
(1,2,4,5)(4,5)
(1,2,4,5)(2,4)

i'm not sure if these are correct because based on directly above i could get a whole lot more than 12
i'm not sure how to answer this
 
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Your first 3 permutations are all the same, so they don't count as 3 distinct answers. In fact, they are all equal to alpha (which obviously commutes with itself). Now, the identity permutation is one answer. But (1 5)(1 4)(1 2), which is one of the other solutions you gave, is exactly (1 2 4 5). Now (1 2 4 5)(2 1) = (1 4 5), which does not commute with (1 2 4 5). In fact, the other two answers you gave also don't work, i.e.(1 2 4 5)(4 5) = (1 2 4) and (1 2 4 5)(2 4) = (1 2 5) do not commute with (1 2 4 5).

So, so far, the only correct answer you've given is [itex]\alpha[/itex], and I've mentioned that you can include [itex]e[/itex] (the identity permuation). There should be some very easy ones you can get, since this is after all [itex]S_6[/itex]. You've got (3 6), and so (1 2 4 5)(3 6) as well. There's 4. See if you can get the other 8. Use the fact that you can compose permuatations with [itex]\alpha[/itex] and (3 6) to get a permutation that commutes with [itex]\alpha[/itex], but make sure to check that the new composition is not equal to one of the one's you've already listed.
 
i was hinking that they had to be equal, which is why i wrote what i wrote, and i also thought 3 and 6 were fixed, but i guess not. I apparently do no understand what exactly commuting means in this situation. My book mentions nothing except that multiplication of disjoint cycles is commutative. I am more confused and frustrated.
 
as update to the original, it turns out there is only 8 so i don't know how anybody could have figured 12. I'm going to search online for help on this and get back with my answer
 
Simply, two permutations a and b commute iff ab = ba.
 

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