Isomorphic groups of integer matrices

My dudes, buckle up, because we're exploring the wonderful realm of abstract algebra again. I found the following problem in the latest number of the 'Gazeta Matematica' by a teacher called Dan Moldovan. I've got to say, the problem is amazing, and I thank this professor for proposing it.

Let $m,n,p>2$ be natural numbers and let $G=[A_1, A_2, ... A_p]$ matrices in $M_{n}(Z)$. To every matrix in G, we attach the reduced matrix modulo m, $\hat A_i$, so that, if $A_i=(a_{i,j})$ is out matrix, then the reduced matrix is $\hat A_i≡(\hat a_{i,j}$, where $\hat a_{i,j}=a_{i,j}$(modulo m). Prove that $\hat G=[\hat A_1, \hat A_2,...\hat A_p]$, together with multiplication in $M_n(Z)$, forms a group that's isomorphic with G.

Solution:

First of all, let us observe that the neutral element of G isn't $I_n$. I know this sounds counter-intuitive, but take the group formed by an idempotent matrix with multiplication. The neutral element there is the matrix itself, so any idempotent matrix can be the neutral element of this group. For the sake of the problem, we shall consider the neutral element of G as $I$. $I$ has the property that $I^2=I$, and $AI=IA=A$ for any element $A$ in G.

Let us consider the case in which $I=O_n$. In this case, $AO_n=A=O_n$, so G={$O_n$}, and $\hat G=[\hat O_n]$, so the two groups are isomorphic.

If $I$ is different from $O_n$, then we shall prove that $gcd(I)=1$. Let $x=gcd(I)$, and it follows that $I=xB$, where B is another matrix in $M_n(Z)$. Since $I^2=I$, $x^2B^2=xB$, so $xB^2=B$. Since $B^2$ is different than 0 (because otherwise $I=O_n$), x divides all elements in B, so $x^2$ divides all elements of I, which implies that $x=1$.

I found the following lemma in the excellent article 'Finite Groups of Matrices Whose Entries Are Integers', published in AMC in 2002 by Mr. James Kuzmanovich and Mr. Andrey Pavlichenkov.(https://www.jstor.org/stable/2695329)

Lemma: Let G be a group of integer matrices in which there is an element $A$ such that $A^q=I$, with q prime, and $\hat A=\hat I$, where $\hat A$ is the reduced matrix modulo p, where p is another prime number. It follows that $A=I$.

Proof: Since $\hat A=\hat I$, $A=I+pH_1$. Let $gcd(H_1)=d$
⟹$A=I+pdH$, where $gcd(H)=1$.
⟹$A^q=(I+pdH)^q=I+\sum_{k=1}^q {{q}\choose{k}}(pd)^{k}H^k I^{q-k}=I$. However, $I^n=I$, for all n, so $\sum_{k=1}^q {{q}\choose {k}}(pd)^{k-1}H^k I=O_n$ (I divided by $pd$. The only element without $p$ in it is is $qHI$, so p divides q, and, thus, p is q or p is 1. Since p is prime, p is q, and, by dividing the relation once more, we get that p divides $gcdH$, so p is 1, which is a contradiction. Thus, $H=O_n$, so $A=I$.                           

Back to our problem, I shall prove something that probably everyone saw coming: the fact that $f(X)=\hat X$ is the isomorphism we were looking for, where $\hat A$ is the reduced matrix modulo m. Obviously, $\hat I$ is the neutral element in $\hat G$, and all the calculations are the same, so f is a morphism. Since $G$ and $\hat G$ are finite, we have to prove that the morphism is injective. For that, we shall prove that $Ker (f)=[I]$.

Let us presume by absurd that there is an A in G such that $f(A)=\hat A=\hat I$. Thus, $A=I+mB$, so, because G has order p, $(I+mB)^p$. Let q be a prime factor of p. Thus, we have that
$$⟹((I+mB)^\frac{p}{q})^q=I$$
Since $\frac {p}{q}=k$ is an integer, $(I+mB)^k=I+mH_1$, so $(I+mH_1)^q=I$.
$$⟹\sum_{j=1}^q {{q}\choose{k}}(mH_1)^j I=O_n$$
Let $gcd H_1=d$, so $H_1=dH$, with $gcd(H)=1$. Thus,
$$\sum_{j=1}^q (md)^{j-1}H^j I= \sum_{j=1}^q (md)^{j-1} H^j =O_n$$
By using the same logic as before, $md | qH$. Since $gcd(H)=1$, m divides q, and, since m is bigger than 1, m is q. Because both m and q are prime, we can use the lemma, and, thus, H=0, so $(I+mB)^k=I$.

By applying this algorithm to all the prime divisors of k, we end up with $mB=O_n$, so $B=O_n$, so $A=I$. Thus, Ker(f)={I}, so the morphism f is injective, and, consequently, bijective, so $G$ and $\hat G$ are isomorphic. This concludes the proof.

I can't really think of a paragraph ending this article, which is a total shame, since this problem was magnificent. I have an excuse though: I have the MAT in a few days, and I am a bit stressed out. I think I'll go play some World of Warcraft to relax. After the MAT, I will try posting more problems, though I came to realize the problems I post while under pressure are longer and harder than the normal problems I post. Maybe it's some kind of psychological mechanism behind this... I'm not foreshadowing the psychology article, btw.

Cris.