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Meepi

  • one year ago

Let \(k\) be a positive integer with \(k \leq |S_n|\). The symmetric group \(S_n\) acts on the set B consisting of all k-element subsets of \({1, \ldots , n}\) by \(\sigma \cdot {a_1, \ldots, a_k}={\sigma(a_1),\ldots,\sigma(a_k)}\). Determine for which values of \(k\) the actions of \(S_n\) on k-element subsets is faithul

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  1. Meepi
    • one year ago
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    Let k be a positive integer with \(k \leq n\). The symmetric group \(S_n\) acts on the set \(B\) consisting of all \(k\)-element subsets of \(\{1,…,n\}\) by \(\sigma⋅\{a_1,\ldots,a_k\}=\{σ(a1),\ldots,σ(ak)\}\). Determine for which values of \(k\) the actions of \(S_n\) on \(k\)-element subsets is faithul

  2. Meepi
    • one year ago
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    Egh, I can't type: \(\{\sigma(a_1), \ldots , \sigma(a_k)\}\)

  3. wio
    • one year ago
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    \(S_n\) is a group and it's operation is symmetric?

  4. Meepi
    • one year ago
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    \(S_n\) is the symmetric group of the set \(\{1, \ldots, n\}\)

  5. Meepi
    • one year ago
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    I know that if k = n, the operation is not faithful since any permutation will still give the n-element subset, and that if k = 1 it i's trivially faithful, but I'm stuck on proving whether or not it's faithful for 1 < k < n

  6. wio
    • one year ago
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    How would it not be faithful?

  7. Meepi
    • one year ago
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    An action is faithful if different elements of \(S_n\) induce different permutations of B, and if k = n, there is only one n-element subset, which every element of S_n maps to. ie if for different \(g, h \in S_n\), \(g\cdot x\neq h\cdot x\) for some \(x \in B\)

  8. wio
    • one year ago
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    Hmmm, one thing I can say is that the set of all subsets is the power set and it has cardinality \(2^k\)

  9. wio
    • one year ago
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    So is \(S_n\) full of \(n\) mappings which will permute the elements in \(B\)?

  10. Meepi
    • one year ago
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    \(S_n\) is the set of n! distinct permutations of the set \(\{1, 2, ..., n\}\) it acts on B by permuting it's elements ie: if we have the permutation (1 2 3) (1 gets send to 2, 2 to 3, 3 to 1) act on {1, 2} we get \((1\,\,2\,\,\,3) \cdot \{1,2\}=\{2, 3\}\)

  11. wio
    • one year ago
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    Should it result in \(\{3,1\}\)?

  12. Meepi
    • one year ago
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    the action is defined as \(\sigma \cdot \{a_1, ... , a_k\} = \{\sigma(a_1), ... ,\sigma(a_k)\}\) so in the case that \(\sigma =\)(1 2 3), \(\sigma \cdot \{1, 2\}=\{\sigma(1), \sigma(2)\}=\{2, 3\}\)

  13. wio
    • one year ago
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    Okay, I think I get that much now.

  14. wio
    • one year ago
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    I've seen this before, but the notation we use for permutation was maybe backwards.

  15. wio
    • one year ago
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    The number of \(k\) element subsets in \(n\) is \(\binom nk\)

  16. wio
    • one year ago
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    It's faithful if no two permutations produce equivalent sets?

  17. wio
    • one year ago
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    Consider what happens with some concrete examples or concrete permutations.

  18. Meepi
    • one year ago
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    now if k =1, there exists two permutations \(\sigma\) and \(\tau\) such that \(\sigma(a) \neq \tau(a)\) if we let \(x = \{a\}\) (which is a subset of {1, ..., n} of cardinality 1) then without loss of generality we have \(\sigma \cdot x \neq \tau \cdot x\), so action is trivial if k is one, if k = n, then there is only one subset of {1, ..., n} namely that set itself so every permutation will give you that subset. now I think for 1 < k < n I will have to show that the permutation representation is injective. I think if we let \(\sigma, \tau\) be different permutations again, then without loss of generality \(\sigma(a_1) = x, \tau(a_2) = x\) then we can choose a subset z of cardinality k so that x and \(a_1\) are in the subset, but \(a_2\) isn't (since k < n, we have at least 1 element not in the subset), then \(\sigma \cdot z \neq \tau \cdot z\), since \(\sigma \cdot z\) contains x, and \(\tau \cdot z\) doesn't, so the action is faithful if k<n

  19. Meepi
    • one year ago
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    I think this works :D

  20. wio
    • one year ago
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    There only needs to be two distinct permutations which don't result in the same subset?

  21. Meepi
    • one year ago
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    No, for any two distinct permutations there must be a subset so that \(\sigma \cdot x \neq \tau \cdot x\). In this proof i think i managed to show that we can always construct at least 1 subset of length k which does this

  22. wio
    • one year ago
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    So if I get this right: If the permutations are distinct, there has to be an output where the permutations have a different input to output it. You then find a subset which has the input for one of the permutations but lacks the needed input for the second permutation. Thus there is an output which one has and the other doesn't.

  23. Meepi
    • one year ago
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    yes

  24. Meepi
    • one year ago
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    and such a subset always exists, since there is at least 1 element from {1, ..., n} not in the subset

  25. wio
    • one year ago
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    You can ensure an excluded input is excluded, and you can also ensure the necessarily included input gets included?

  26. Meepi
    • one year ago
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    Yes, since we only consider 2 elements, a1 and x, and for 1 < k < n, k is at least 2 :)

  27. wio
    • one year ago
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    What if a1 and a2 are greater than k, does that matter?

  28. Meepi
    • one year ago
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    Not really, since k is just how much elements are in the subset

  29. Meepi
    • one year ago
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    ie if we have k = 2, and two permutations \(\sigma(7) = 8\) and \(\tau(8)=8\) then we choose a subset which doesn't have 7 in it, but contains 8 and any other element, say 5 so we get {5, 8} now even if the 2 permutations map 5 to the same element, 8 still gets mapped to a different one

  30. Meepi
    • one year ago
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    two permutations such that*

  31. Meepi
    • one year ago
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    Thanks for the help by the way, trying to explain it to someone else always helps it seems :)

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