# Permutation 1

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 dont 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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AKG
Homework Helper
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 $\alpha$, and I've mentioned that you can include $e$ (the identity permuation). There should be some very easy ones you can get, since this is after all $S_6$. 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 $\alpha$ and (3 6) to get a permutation that commutes with $\alpha$, 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 gonna search online for help on this and get back with my answer

AKG