The group algebra $\mathbb{C}S_n$ has a simple module of dimension $n-1$ over $\mathbb{C}$. A hint has been given here but I didn't quite understand it + the question is 2 years old, so I decided to ask a new question.
So $A = \mathbb{C}S_n$ will be the group algebra that is the $\mathbb{C}$-span of $\{ v_g : g\in S_n \}.$ In the comments of the original post, it says that the span $V$ of $$\{(1,0,\ldots,0 ,-1),(0,1,\ldots ,0,-1),\ldots ,(0,0,\ldots,1,-1)\}\subset\mathbb{C}^{n}$$ is a simple $\mathbb{C}S_{n-1}$-module, but I am not even sure how the algebra actually acts on the module $V$, let alone show that it is simple. Does $v_g$ act on $V$ by switching the basis elements according to $g$, or does it switch the components of the vectors in $V$ (for example $v_{(12)}.v_1,v_2,v_3 = (v_2,v_1,v_3)$)? Any help would be appreciated.
 A: $S_n$ acts on $\mathbb{C}^n$ by permutation, and there are two conventions which work (permute basis vectors or permute components), but permuting basis vectors ends up being a bit more natural.  Here is a little background first.  Given a group $G$ and two sets $A$ and $B$ which $G$ acts on, then the set of functions $A\to B$ is acted upon by $G$ using the action defined by $(gf)(a)=gf(g^{-1}a)$.  The inverse makes it so $g(hf)=(gh)f$.
A vector is a function $v:[n]\to \mathbb{C}$, and viewing $[n]=\{1,2,\dots,n\}$ as a set with an $S_n$ action (in particular, the defining action of $S_n$) and viewing $\mathbb{C}$ as the trivial representation, $S_n$ acts on vectors by $(\sigma v)_i=v_{\sigma^{-1}(i)}$.  On standard basis vectors, one can verify that $\sigma e_i=e_{\sigma(i)}$:
$$(\sigma e_i)_j=\delta_{i,\sigma^{-1}(j)}=\delta_{\sigma(i),j}=(e_{\sigma(i)})_j$$
Let $V$ be the span of the vectors you describe, so $V$ is the orthogonal complement of the vector $(1,1,\dots,1)$, or in other words the set of vectors whose components sum to $0$.  Certainly, permuting the entries of a vector will not change whether the components sum to $0$, so $V$ is a subrepresentation of dimension $n-1$.
That $V$ is simple can be shown by showing that it is cyclic for any nonzero $v\in V$.  Just give a process which can take an arbitrary vector and give one of the basis vectors through a sequence of permuting entries and linear combinations, then show that every basis vector can be reached from that basis vector.
Another way is to calculate the character $\chi$ of the representation and show that $(\chi,\chi)=1$.  (Easiest by starting with the character of the $\mathbb{C}^n$ representation and subtracting off the trivial representation.)
