Mistake in proof: Symmetric group acting on group ring Let $G$ be finite group of order $n$ and $K$ be a field. The symmetric group on $G$, denoted by $S_G$ acts linearly on the group ring $K[G]$ by
$$\sigma\left(\sum_{g\in G}c_gg\right):=\sum_{g \in G}c_g\sigma(g) $$
The corresponding homomorphism
$$\rho:S_G\to GL(K[G])\cong GL_n(K)  $$
is, injective since $\rho(\sigma)=id$ implies $\sigma(g)=g$ for all $g\in G$.
Now if $K=\mathbb{F}_p$ this would imply that $n!=\# S_G$ divides $\# GL(K[G])=\#GL_n( \mathbb{F}_p)$, which is absolute nonsense since I didn't use that $G$ is a $p$-group or anything similar. Where did I make a mistake?
 A: There is no mistake and as JBL says in the comments you don't use anything about the group structure on $G$. Here is a simpler argument: $S_n$ acts $K$-linearly and faithfully on $K^n$ by permuting the coordinates, for any field $K$ (even any commutative ring). This gives a canonical embedding of $S_n$ into $GL_n(K)$ as the subgroup of permutation matrices which, when $K$ is a finite field $\mathbb{F}_q$, establishes that $n!$ divides the order
$$|GL_n(\mathbb{F}_q)| = q^{ {n \choose 2} } (q - 1)^n [n]_q!$$
of $GL_n(\mathbb{F}_q)$. Here
$$[n]_q! = \prod_{i=1}^n [i]_q = \prod_{i=1}^n \frac{q^i - 1}{q - 1}$$
is the $q$-factorial. It is by no means obvious at first glance, just looking at these formulas, that this divisibility ought to hold. So, how can we prove it directly? We can work one prime at a time. If $p$ is a prime recall that $\nu_p(n)$ is the greatest exponent $k$ such that $p^k \mid n$. Legendre's formula tells us that
$$\nu_p(n!) = \sum_{k \ge 1} \left\lfloor \frac{n}{p^k} \right\rfloor$$
so to prove the divisibility we need to show that $\nu_p(|GL_n(\mathbb{F}_q)|)$ is bigger than or equal to this for all primes $p$. I did this calculation recently here so we can steal it: if $\gcd(p, q) = 1$ then we get
$$\nu_p(|GL_n(\mathbb{F}_q)|) = \left\lfloor \frac{n}{d} \right\rfloor \nu_p(q^d - 1) + \sum_{k \ge 1} \left\lfloor \frac{n}{dp^k} \right\rfloor$$
where $d = \text{ord}_p(q)$ is the smallest positive integer such that $q^d \equiv 1 \bmod p$. In particular $d \le p-1$ and $\nu_p(q^d - 1) \ge 1$ which gives
$$\nu_p(|GL_n(\mathbb{F}_q)|) \ge \sum_{k \ge 0} \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor$$
and the desired inequality is clear since $\frac{1}{p-1} > \frac{1}{p}$. If $p \mid q$ then $\nu_p(|GL_n(\mathbb{F}_q)|) \ge {n \choose 2}$ which is much larger than what we need.
This expression on the RHS can be interpreted as $\log_p$ of the order of the "Sylow $p$-subgroup" of $GL_n(\mathbb{Q})$, in the sense that every $p$-subgroup of $GL_n(\mathbb{Q})$ embeds into it (which means it controls how large a finite subgroup of $GL_n(\mathbb{Q})$ can be). See Serre's Bounds for the orders of the finite subgroups of $G(k)$ for much more on this.
