# Find the least interger m such that $S_n$ is embedd into $GL_m(F)$

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 $GL_m(F)$?

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$V_n=\{x\in F^m\ | \ \sum x_i=0\}$ (with $S_n$ permuting the $x_i$) is a faithful representation of minimal dimension for $\text{char}(F)\not|n$, and $V_n/\langle(1,...,1)\rangle$ for $\text{char}(F)|n$ – yoyo Sep 4 '12 at 23:22

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.

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@Holt, could you give more details? – ougao Sep 4 '12 at 22:26
@ougao: I think he already did give you the full recipe as well as a reference to a proof. Your question properly falls into the area of [tag: representation-theory], so you may want to study that (unless you already have). To find the matrices all you need to do is to fix a basis of $U$ (or $U/V$ when $p\mid n$) and start computing, how the various permutations act on the basis vectors. – Jyrki Lahtonen Sep 5 '12 at 8:09
In fact I added most of this after his request for more details, so the request was reasonable! – Derek Holt Sep 5 '12 at 9:01
@Holt, thanks for the detail and reference – ougao Sep 5 '12 at 15:38
Ok, sorry. Should have checked the time stamps and the edit history. – Jyrki Lahtonen Sep 5 '12 at 20:58