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Meepi
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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
 9 months ago
 9 months ago
Meepi Group Title
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
 9 months ago
 9 months ago

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Meepi Group TitleBest ResponseYou've already chosen the best response.1
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
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
Egh, I can't type: \(\{\sigma(a_1), \ldots , \sigma(a_k)\}\)
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
\(S_n\) is a group and it's operation is symmetric?
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
\(S_n\) is the symmetric group of the set \(\{1, \ldots, n\}\)
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
I 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
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
How would it not be faithful?
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
An 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 \(x \in B\)
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Hmmm, one thing I can say is that the set of all subsets is the power set and it has cardinality \(2^k\)
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
So is \(S_n\) full of \(n\) mappings which will permute the elements in \(B\)?
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
\(S_n\) 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 \((1\,\,2\,\,\,3) \cdot \{1,2\}=\{2, 3\}\)
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Should it result in \(\{3,1\}\)?
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
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), \(\sigma \cdot \{1, 2\}=\{\sigma(1), \sigma(2)\}=\{2, 3\}\)
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Okay, I think I get that much now.
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
I've seen this before, but the notation we use for permutation was maybe backwards.
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
The number of \(k\) element subsets in \(n\) is \(\binom nk\)
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
It's faithful if no two permutations produce equivalent sets?
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Consider what happens with some concrete examples or concrete permutations.
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
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
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
I think this works :D
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
There only needs to be two distinct permutations which don't result in the same subset?
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
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
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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.
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
and such a subset always exists, since there is at least 1 element from {1, ..., n} not in the subset
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
You can ensure an excluded input is excluded, and you can also ensure the necessarily included input gets included?
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
Yes, since we only consider 2 elements, a1 and x, and for 1 < k < n, k is at least 2 :)
 9 months ago

wio Group TitleBest ResponseYou've already chosen the best response.1
What if a1 and a2 are greater than k, does that matter?
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
Not really, since k is just how much elements are in the subset
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
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
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
two permutations such that*
 9 months ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
Thanks for the help by the way, trying to explain it to someone else always helps it seems :)
 9 months ago
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