For which values of $k$ the action of $S_n$ on $k$-element subsets is faithful?

Question

Suppose $S_n$ acts on a set of $k$ element subsets of $\{1,\ldots,n\}$, by $\sigma\cdot\{a_1,\ldots,a_k\}=\{\sigma(a_1),\ldots,\sigma(a_k)\}$ where $k<n$ and $\sigma\in S_n$. Show that action is faithful.

Proof: Suppose action not faithful, then there exist $\sigma, \tau\in S_n$ such that $\sigma\cdot\{a_1,\ldots,a_k\}=\tau\cdot\{a_1,\ldots,a_k\}$ for all $k$ element subsets $\{a_1,\ldots,a_k\}$, but $\sigma\ne \tau$. Then there exists an $x\in\{1,\ldots,n\}$ such that $\sigma(x)\ne\tau(x)$. But I can't see how to show $\sigma\cdot\{x,\ldots,a_k\}\ne\tau\cdot\{x,\ldots,a_k\}$.

Answer 1

You need to assume that $1 \le k$ as well as $k<n$.

Suppose that $1 \ne \sigma \in S_n$. So there exists $x$ with $\sigma(x) \ne x$. Then $\sigma^{-1}(x) \ne x$, so there exists a $k$-subset $S$ with $x \in S$ but $\sigma^{-1}(x) \not\in S$. Then $x \not\in \sigma(S)$, so $\sigma(S) \ne S$. So the action of $S_n$ on the set of $k$-subsets is faithful.

Answer 2

Let $X_k:=\{S\subseteq \{1,\dots,n\}\mid \#S=k\}$. Let's define, $\forall \sigma\in S_n, \forall S\in X_k$:

$$\sigma\cdot S:=\sigma(S) \tag 1$$

Since $\#\sigma(S)=k$, $(1)$ defines an action of $S_n$ on the set $X_k$. The kernel of this action is:

\begin{alignat}{1} \bigcap_{S\in X_k}\operatorname{Stab}(S) &= \bigcap_{S\in X_k}\{\sigma\in S_n\mid \sigma(S)=S\} \\ &= \{\sigma\in S_n\mid \sigma(S)=S, \forall S\in X_k\} \\ &= \{\iota\} \end{alignat}

In fact, if $\sigma \ne \iota$, then $\exists S\in X_k$ such that $\sigma(S)\ne S$ (see the argumentation in Derek Holt's answer).

Therefore, the kernel of the action is trivial and the action is faithful.