Does $S_n$ always embed into $GL_{n-1} (\mathbb{F}_p$)?

Question

$S_n$ is the symmetry group of the standard $n-1$-simplex, which is the convex hull of the standard basis vectors in $\mathbb{R}^n$. One can orthogonally project this shape onto the plane $x_1 +...+ x_n = 0$, which is isomorphic to $\mathbb{R}^{n-1}$. This yields an embedding of $S_n$ into $GL_{n-1}(\mathbb{R}$).

Is there some other trick one can use for more general fields, like prime fields? It seems likely, because $GL_{2} (\mathbb{F}_2)$ is isomorphic to $S_3$, and $GL_{3} (\mathbb{F}_2$) contains a copy of $S_4$ according to: https://people.maths.bris.ac.uk/~matyd/GroupNames/163/GL(3,2).html. And as $n$ grows large the order of $GL_{n-1} (\mathbb{F}_p)$ vastly exceeds the order of $S_n$. I have no idea how one would go about proving it though.

Answer 1

Okay, I was being a little flippant. There are actually some nontrivial details to check here because of positive characteristic issues so let's check them.

Let $F$ be any field. $S_n$ acts faithfully by permutation on $F^n$, producing a natural embedding $S_n \hookrightarrow GL_n(F)$. To reduce the value of $n$ by $1$ we actually have two options:

• First we can observe that the action of $S_n$ on $F^n$ has a $1$-dimensional invariant subspace spanned by $(1, 1, \dots )$ and we can quotient by this subspace. This gives us an $(n-1)$-dimensional representation which is a quotient of $F^n$; call it $V$.
• Or we can observe that there is a natural map $F^n \to F$ given by adding up all the coordinates which is $S_n$-invariant (where $F$ is given the trivial $S_n$-action), and we can take its kernel. This gives us another $(n-1)$-dimensional representation which is a subspace of $F^n$; call it $W = \{ (x_1, \dots x_n) \in F^n : \sum x_i = 0 \}$.

If $\text{char}(F) \nmid n$, then these representations are isomorphic. This is because in this case the trivial subrepresentation $F = \text{span}((1, 1, \dots 1))$ of $F^n$ is a direct summand, because we can project onto it via the averaging map

$$P : F^n \ni (x_1, \dots x_n) \mapsto \frac{x_1 + \dots + x_n}{n} (1, 1, \dots 1).$$

This gives us a direct sum decomposition $F^n \cong \text{im}(P) \oplus \text{ker}(P)$ (this is true more generally for any idempotent) and the other summand in that decomposition is identified with both $V$ and $W$. Since $F^n$ is a faithful action of $S_n$ and the action on $F$ is trivial, this implies that both $V$ and $W$ are faithful, so we get an embedding $S_n \hookrightarrow GL_{n-1}(F)$ using either of them, as desired.

However, if $\text{char}(F) \mid n$ then this averaging map is no longer defined, and we run into problems with $W$. The issue is that $\sum_{i=1}^n 1 = n = 0$ in this case, so $W$ actually contains the trivial representation spanned by $(1, 1, \dots 1)$. In fact if $\text{char}(F) = n = 2$ we see that $W$ is exactly this trivial representation and in particular is not faithful, and actually neither is $V$.

So we must work with the quotient $V$ rather than the subspace $W$, and now it's not entirely clear that the action remains faithful so we need to check this. Explicitly we need to check that if $\pi \in S_n$ is a non-identity permutation then there exists a vector $x = (x_1, \dots x_n)$ such that the permuted vector $\pi x = (x_{\pi^{-1}(1)}, \dots x_{\pi^{-1}(n)})$ does not differ from $x$ by a multiple of $(1, 1, \dots 1)$. Consider any cycle in $\pi$, which WLOG we can take to be the cycle $(123 \dots k)$ for some $k$. If we take $x = (1, 0, 0, \dots 0)$ to be an element of the standard basis then $\pi x = (0, 1, 0, \dots 0)$ differs from $x$ by $(-1, 1, 0, \dots 0)$ which is not a multiple of $(1, 1, 1, \dots )$ unless $\text{char}(F) = n = 2$ as above.

So, the conclusion is that we can embed $S_n$ into $GL_{n-1}(F)$ for any field $F$ using the quotient $V$ above unless $\text{char}(F) = n = 2$. In this case it's not hard to see that $S_2 \cong C_2$ doesn't embed into $GL_1(F) \cong F^{\times}$ because the polynomial $x^2 - 1 = (x - 1)^2$ has no nontrivial roots in characteristic $2$.

Answer 2

This answer complements the other one by showing via a different argument that the action on $W$ is also faithful (this does not contradict $W$ containing the trivial representation in certain cases).

First we observe that $S_2$ does not embed in $\mathrm{GL}_1(\mathbf{F}_2)=\{1 \}$ but does embed in $\mathrm{GL}_1(\mathbf{F}_p) \cong \mathbf{Z}/(p-1) \mathbf{Z}$ for all odd primes $p$.

Suppose then that $n \geq 3$. Check that given a non-trivial permutation $w \in S_n$, there always exists a two element subset $\{i,j\}$ of $\{1,2,\dots,n\}$ that is not fixed by $w$. For instance, this follows by noting that otherwise the cycle decomposition of $w$ involves only $1$-cycles and $2$-cycles, hence must involve at least two different cycles at least one of which is non-trivial, and choosing an element from each gives a non-stable subset.

This implies that the permutation action of $S_n$ on the set $$R=\{\epsilon_i-\epsilon_j \ | \ 1 \leq i\neq j \leq n\} \subseteq W$$ is faithful (even when $p=2$), where $\epsilon_i$ is the vector with a $1$ in position $i$ and $0$'s elsewhere, and a fortiori the $S_n$ action on $$W=\{x_1+x_2+\dots+x_n=0 \} \subseteq \mathbf{F}_p^n$$ is faithful.