Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.10
Let $k \in \mathbb{Z}^+$.
(1) For which values of $k$ is the action of $S_A$ on $k$-element subsets faithful?
(2) For which values of $k$ is the action of $S_A$ on ordered $k$-tuples faithful?
Solution:
(1) First assume that $k < |A|$, and suppose $\sigma, \tau \in S_n$ such that $$\sigma \cdot \{ a_i \}_{i = 1}^k = \tau \cdot \{ a_i \}_{i = 1}^k$$ for all $k$-element subsets $\{ a_i \}_{i = 1}^k$, but that $\sigma \neq \tau$. Apply $\tau^{-1}$ to both sides, we may assume that $\tau=\mathsf{id}$. We would like to get a contradiction. Suppose $\alpha$ is a positive integer such that $\sigma(\alpha)\ne \alpha$. Then take any $k$-element subset $\{ a_i \}_{i = 1}^k$ excluding $\sigma(\alpha)$, then we have $\sigma(\alpha)\in \sigma \cdot \{ a_i \}_{i = 1}^k $ but $\sigma(\alpha)\notin \{ a_i \}_{i = 1}^k $. Hence $$\sigma \cdot \{ a_i \}_{i = 1}^k = \tau \cdot \{ a_i \}_{i = 1}^k$$ does not hold for this particular $k$-subset. Therefore such $\sigma$ can only be $\mathsf{id}$. Namely in this case, the action on $k$-element subsets is faithful.
If $k=|A|$, then it is clear that this action is not faithful unless $|S_A|=1$. In the case $|A| = 1$, we must have $k = 1$. The action is faithful in a trivial way; both $S_A$ and the set of all 1-element subsets of $A$ have only one element.
In summary, if $|A| = 1$ the action is faithful. If $|A| > 1$, the action is faithful if and only if $k < |A|$.
(2) Suppose $\sigma, \tau \in S_n$ such that $$\sigma \cdot ( a_i )_{i = 1}^k = \tau\cdot ( a_i )_{i = 1}^k$$ for all $k$-tuples $( a_i )_{i = 1}^k$. By the definition of equality for $k$-tuples, and because every element of $A$ appears in at least one tuple, we have $\sigma(a) = \tau(a)$ for each $a \in A$. Thus $\sigma = \tau$, and so the action is faithful for all $k \in \mathbb{Z}^+$.
