TITLE: Find the least interger m such that $S_n$ is embedd into $GL_m(F)$
QUESTION [8 upvotes]: let $S_n$ be the permutation group on n letters, we know that there exists an injective group homomorphism from $S_n$ to $GL_m(F)$, where $F$ denotes a field, if $m=n$, so my question is:
For a fixed n, what is the least possible nonzero integer of m, such that $S_n$ can be embeded into $GL_m(F)$?

REPLY [4 votes]: If $F$ has characteristic 0, then the least such $m$ is $n-1$.
The same is true if $F$ has characteristic $p$ unless $p|n$, in which case it is $n-2$ (for $n>4$).
The standard permutation representation of $S_n$, in which elements of $S_n$ are represented by their corresponding permutation matrices, can be defined over any field and has degree $n$.
In characteristic zero or when $p$ does not divide $n$, the corresponding $FS_n$-module decomposes as a direct sum of modules $U$ of dimension 1, spanned by the vector $(1,1,\ldots,1)$, and $V$ of dimension $n-1$ spanned by the vectors with coefficients  summing to 0 (as defined in Yoyo's comment). It can be proved that $V$ is irreducible.
When $p|n$, we have $U < V$, and $V/U$ is irreducible of dimension $n-1$.
It can be proved that these are the smallest degree faithful representations of $S_n$ (at least for $n>4$ - in the modular case, they are not always faithful when $n \le 4$).
A good reference for this topic is:
James, G. D. (1983), "On the minimal dimensions of irreducible representations of symmetric groups", Mathematical Proceedings of the Cambridge Philosophical Society 94 (3): 417–424.