# Permutations of $k$-element Subsets

Let $A$ be a nonempty subset, $k$ a positive integer such that $k \le |A|$, and $B = \{S \subseteq A ~|~ |S| = k \}$. Consider the action of $S_A$ on $B$ defined by $\sigma \cdot \{a_1,...,a_k\} = \{\sigma(a_1),...,\sigma(a_k)\}$. For what $k$ is this action faithful?

My conjecture is that the action is faithful if and only if $k=1$, but evidently this is wrong. Here is my proof:

The $\impliedby$ direction is obvious. To prove the other direction, suppose that the action is faithful yet $k >1$. Because the action faithful, the kernel must be trivial, so that $e$, the identity of $S_A$, is the only permutation for which $e \cdot \{a_1,...,a_k\} = \{a_1,...,a_k\}$ holds for all $k$-element subsets of $A$. However, consider the permutation $\tau$ that takes $a_1$ to $a_2$ and vice-versa. Then clearly $\tau$ is not the identity yet it takes $\{a_1,...,a_k\}$ to itself.

I think the issue is that $\tau$ takes $\{a_1,...,a_k\}$ to itself, but that is only one $k$-element subset of $A$. Does this appear to be my issue?

Let it be that $1\leq k<|A|$.
If $\tau\neq e$ then $a\neq \tau(a)$ for some $a \in A$.
Now some $S\in B$ can be chosen with $a\in S\wedge \tau(a)\notin S$.
But we do have $\tau(a)\in\tau.S$ so that $\tau.S\neq S=e.S$.