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Egh, I can't type: \(\{\sigma(a_1), \ldots , \sigma(a_k)\}\)

\(S_n\) is a group and it's operation is symmetric?

\(S_n\) is the symmetric group of the set \(\{1, \ldots, n\}\)

How would it not be faithful?

So is \(S_n\) full of \(n\) mappings which will permute the elements in \(B\)?

Should it result in \(\{3,1\}\)?

Okay, I think I get that much now.

I've seen this before, but the notation we use for permutation was maybe backwards.

The number of \(k\) element subsets in \(n\) is \(\binom nk\)

It's faithful if no two permutations produce equivalent sets?

Consider what happens with some concrete examples or concrete permutations.

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

I think this works :D

There only needs to be two distinct permutations which don't result in the same subset?

yes

Yes, since we only consider 2 elements, a1 and x, and for 1 < k < n, k is at least 2 :)

What if a1 and a2 are greater than k, does that matter?

Not really, since k is just how much elements are in the subset

two permutations such that*

Thanks for the help by the way, trying to explain it to someone else always helps it seems :)