Proving an action is faithful My question is this; Prove that the action of $S_n$ on $Y_k$ and $Y_{k} =$ $\dbinom{X}{k}$, where $X=\{1,2,...,n\}$ is faithful for all $1< k < n$. 
I know that any action is just fundamentally a permutation therefore we have that all actions are injective, but I am having difficulty trying to express this idea in a rigorous way. 
 A: A good strategy is often just write out exactly what every statement you are given or want to prove means.  You want to prove the action of $S_n$ on $Y_k$ is faithful.  This means that if $\sigma\in S_n$ is not the identity, then it does not act by the identity: that is, there is some $A\in Y_k$ such that $\sigma(A)\neq A$.
So suppose you have $\sigma\in S_n$ which is not the identity.  You want to find some $k$-element subset $A\subset X$ such that $\sigma(A)\neq A$.  You're going to need to use the fact that $\sigma$ is not the identity, which mean that there is some $i\in X$ such that $\sigma(i)\neq i$.  So maybe we should try using this $i$ to find our set $A$.  For instance, suppose we wanted to pick a set $A$ such that $i\in A$.  How should we pick the remaining $k-1$ elements of $A$ to be sure that $\sigma(A)$ will not be equal to $A$ (using the fact that $\sigma(i)\neq i$)?
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 If we can guarantee that $\sigma(i)\not\in A$, then we'll be done, since $\sigma(i)\in\sigma(A)$.  So we need to pick our remaining $k-1$ elements so that none of them are equal to $\sigma(i)$.  We can do this since $k<n$, so that there are $n-2\geq k-1$ elements in $A$ besides $i$ (which we've already put in $A$) and $\sigma(i)$ (which we don't want to put in $A$).  Note that we don't actually need $k>1$; we just need $k>0$ (so that we are allowed to have $i\in A$ to begin with!).

