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