Problem Statement:
Let $A$ be a nonempty set and let $k$ be a positive integer with $k\leq\left|A\right|$. The symmetric group $S_A$ acts on the set $B$ consisting of all subsets of $A$ of cardinality $k$ by $\sigma\cdot\{a_1,...,a_n\}=\{\sigma(a_1),...\sigma(a_n)\}$.
For which values of $k$ is the action of $S_n$ (the symmetric group) on $k$-element subsets faithful?
My Answer:
Case 1: $k=\left|A\right|=1$
The only permutation group in $S_1$ is $1$ therefore k is faithful on $B$.
Case 2: $k=\left|A\right|\neq1$ and finite
The cardinality of $B$ is 1. It is the set of all elements of $A$. Therefore all permutations mapping $B$ to $B$ are the same and $S$ is not faithful.
Case 3: $k<\left|A\right|\neq1$ and finite
For any $b\in B$ a transposition of any two $a\in b$ is a permutation such that $\sigma b=b$. Therefore $S$ is not faithful.
Discussion:
I've googled up three solution manuals that say that $S$ is faithful in case 3. Each provides a different proof, none of which I can fully track. I think I must be misunderstanding a definition somewhere.
Here is an example that I think illustrates S not being faithful in case 3:
$A=\{1,2,3,4\}$
$b=\{1,2\}$
$\sigma=(1 2)$
$\sigma b=\{2,1\}=\{1,2\}=b$
Where am I wrong? This is self-study, so any additional coaching on my thinking would be appreciated.
What I missed was that if S is not faithful then $\sigma_2^{-1}\sigma_1b=\sigma b=b$ and $\sigma_1\neq\sigma_2\iff\sigma\neq 1$ must hold for all $b\in B$ for some $\sigma\in S$. I thought I could get away with only showing it for a single element $b$ because I did not understand the definition of faithful.
In case 3 for any $\sigma\neq1$ we can choose a $b$ that doesn't contain the second element of the first cycle of $\sigma$ but does contain the first element. This means that $\sigma b\neq b,$ which is a contradiction. We can choose no $\sigma$ meeting the requirements for all $b$. As a result S must be faithful.