Show that the order of a finite subgroup of $\mathrm{SL}_n(\mathbb{Z})$ divides the order of $\mathrm{SL}_n(\mathbb{F_3})$

Question

This problem comes from Johns Hopkins University Spring 2020 algebra qualifying.
Let $G$ be a finite subgroup of $\mathrm{SL}_n(\mathbb{Z})$. Prove that the order of $G$ divides $$ \frac{1}{2}\left(3^n-1\right)\left(3^n-3\right) \ldots\left(3^n-3^{n-1}\right) . $$ Hint: Use reduction modulo 3.

My idea: Assuming $\pi_p:\mathrm{SL}_n(\mathbb{Z}) \rightarrow \mathrm{SL}_n(\mathbb{F_3})$ performed by mod 3 ,so it is sufficient to prove $\pi_p$ provides a injection between finite subgroups of $\mathrm{SL}_n(\mathbb{Z})$ to $\mathrm{SL}_n(\mathbb{F_3})$ , for which I cannot give a proof.
I have found it have been solved on https://en.wikipedia.org/wiki/Congruence_subgroup ,where it provides a huge paper concerning some advanced method. But I havn't learn algebraic number theory. So I wonder some simple method for graduates level students.

Answer 1

Let $g$ be a finite order nontrivial element of the kernel of $SL(n, {\mathbb Z})\to SL(n, {\mathbb Z}_{p})$, $p\ge 3$, prime; I am using the topologist notation for finite abelian groups. (There is nothing special about $p=3$.) Our goal is to reach a contradiction. (The argument is due to Minkowski or to Schur, I am not sure.)

We have $g=I+pA$, where $A\ne {\mathbf 0}$ is an integer matrix. Without loss of generality, we may assume that $g$ has prime order $q$. We have (by applying the binomial formula): $$ g^q= (I+p A)^q= I + {q\choose 1} pA + {q\choose 2} (pA)^2 + {q\choose 3} (pA)^3+... + (pA)^q= I. $$ Factoring out $pA$ (and canceling $p$), we obtain $$ A\underbrace{({q\choose 1} I + {q\choose 2} p A + {q\choose 3} (pA)^2 +... + (pA)^{q-1})}_{B}= {\mathbf 0}. $$
If $B$ is invertible (as a real matrix), we obtain that $A= {\mathbf 0}$, a contradiction. Hence, $B$ has zero determinant. Reducing modulo $p$, we obtain that the determinant of $B$ is $q^n$ modulo $p$. There are now two cases to consider:

(1) $q\ne p$ (recall that $p, q$ are prime). Hence, $\det B$ is nonzero modulo $p$, which is a contradiction.

(2) $q=p$, $$ B= {p\choose 1} I + {p\choose 2} p A + {p\choose 3} (pA)^2 +... + (pA)^{p-1}. $$

Note that if $p=3$, we have $$ B= 3\ I + 3^2 A+ (3A)^2. $$

Observe that each summand in the matrix $B$ is divisible by $p$. Moreover, since $p=q\ge 3$, each summand of $B$, except for the first one, ${p \choose 1} I $, is divisible by $p^2$. Hence, $B= p(I + pC)$ for some integer matrix $C$. Repeating the argument in Case 1, we obtain that $\det(I+pC)=0$ but $\det(I+pC)\ne 0$ modulo $p$. A contradiction.