Matrix of finite order in the kernel of group morphism induced by the reduction morphism $M_n(\mathbb{Z}) \longrightarrow M_n(\mathbb{Z}/2\mathbb{Z})$

Question

I would like to show that any matrix of finite order in the kernel of group morphism induced by the reduction morphism $M_n(\mathbb{Z}) \longrightarrow M_n(\mathbb{Z}/2\mathbb{Z})$ is involutive ($M^2 = I_n$).

Here is what I tried : first I wrote $M = I_n +2dN$ where $d$ is the $gcd$ of coefficients of $\dfrac{1}{2}(M - I_n)$ and assumed that $p$ is the order of $M$, by developing the formula $I_n = (I_n +2dN)^p$ we can show that p is a power of $2$ but I cannot show more despite my efforts.

Any help is welcome.

Answer 1

It is enough to show that if $M^4=I_n$ and $M=I_n+2N$, then $M^2=I_n$.

Write $M=I_n+2N$. If $M^4=I_n$, then substituting $M=I_n+2N$ and dividing by $8$ yields $N+3N^2+4N^3+2N^4=0$, ie $N+N^2+2N^2(N^2+2N+I_n)=0$.

Write $D=N(N+I_n)$, then $M^2-I_n=4D$ and $D+2D^2=0$.

Let $t \geq 0$ be such that $2^t | D$. Then $2^{2t+1} | -2D^2=D$, and $2t+1 > t$.

Thus $D=0$ so $M^2=I_n$.