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anonymous
 2 years 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 kelement 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 kelement subsets is faithul
anonymous
 2 years 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 kelement 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 kelement subsets is faithul

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Let 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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Egh, I can't type: \(\{\sigma(a_1), \ldots , \sigma(a_k)\}\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0\(S_n\) is a group and it's operation is symmetric?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0\(S_n\) is the symmetric group of the set \(\{1, \ldots, n\}\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I know that if k = n, the operation is not faithful since any permutation will still give the nelement 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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0How would it not be faithful?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0An action is faithful if different elements of \(S_n\) induce different permutations of B, and if k = n, there is only one nelement 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 \(x \in B\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Hmmm, one thing I can say is that the set of all subsets is the power set and it has cardinality \(2^k\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0So is \(S_n\) full of \(n\) mappings which will permute the elements in \(B\)?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0\(S_n\) 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 \((1\,\,2\,\,\,3) \cdot \{1,2\}=\{2, 3\}\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Should it result in \(\{3,1\}\)?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0the 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), \(\sigma \cdot \{1, 2\}=\{\sigma(1), \sigma(2)\}=\{2, 3\}\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, I think I get that much now.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I've seen this before, but the notation we use for permutation was maybe backwards.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0The number of \(k\) element subsets in \(n\) is \(\binom nk\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0It's faithful if no two permutations produce equivalent sets?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Consider what happens with some concrete examples or concrete permutations.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I think this works :D

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0There only needs to be two distinct permutations which don't result in the same subset?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0No, 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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0So 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.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0and such a subset always exists, since there is at least 1 element from {1, ..., n} not in the subset

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0You can ensure an excluded input is excluded, and you can also ensure the necessarily included input gets included?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, since we only consider 2 elements, a1 and x, and for 1 < k < n, k is at least 2 :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0What if a1 and a2 are greater than k, does that matter?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Not really, since k is just how much elements are in the subset

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ie 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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0two permutations such that*

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks for the help by the way, trying to explain it to someone else always helps it seems :)
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