Symmetric groups and the "field with one element" I have heard several times that one may regard the symmetric group on $n$ letters as the general linear group in dimension $n$ over the "field with one element". In particular this heuristic would imply, using the Barrat--Priddy--Quillen theorem, that the algebraic $K$-theory groups of the "field with one element" ought to be the stable homotopy groups of spheres. 
Can someone give me some insight as to why the symmetric group is a reasonable choice for the general linear group?
 A: Let's count. There is a well-known formula for the number of ways of choosing an ordered set of $n$ linearly independent 1-dimensional subspaces of an $n$-dimensional vector space over $\mathbb{F}_q$, namely:
$$\frac{q^n - 1}{q - 1} \frac{q^n - q}{q - 1} \cdots \frac{q^n - q^{n-1}}{q - 1} = q^{\frac{1}{2} (n - 1) n} \frac{q^n - 1}{q - 1} \frac{q^{n-1} - 1}{q - 1} \cdots \frac{q - 1}{q - 1}$$
Equivalently, this is the number of flags in an $n$-dimensional vector space over $\mathbb{F}_q$. Now, observe that
$$\frac{q^m - 1}{q - 1} = q^{m-1} + \cdots + q + 1$$
and so, taking the limit $q \to 1$, we find that the number of all "ordered sets of $n$ linearly independent 1-dimensional subspaces of an $n$-dimensional vector space over $\mathbb{F}_1$" is just $n !$. But vector spaces over $\mathbb{F}_1$ are supposed to have no scalar multiplication, so this should also be the number of all "ordered bases for an $n$-dimensional vector space over $\mathbb{F}_1$". But the set of all ordered bases for an $n$-dimensional vector space over $\mathbb{F}_q$ has a canonical free and transitive $\mathrm{GL}_n (\mathbb{F}_q)$-action, so this suggests that $\mathrm{GL}_n (\mathbb{F}_1)$ should be a group of $n !$ elements, such as the symmetric group.
Of course, one could also ask why we don't just look at the formula for the cardinality of $\mathrm{GL}_n (\mathbb{F}_q)$ directly. Well, we could: it's just $(q - 1)^n$ times the number of ordered sets of $n$ linearly independent 1-dimensional subspaces of an $n$-dimensional vector space over $\mathbb{F}_q$. But that obviously goes to $0$ as $q \to 1$, which doesn't make sense for a group. So maybe the above story is just a post facto justification.
