$I_n+pX\in\mathbb{Z}^{n\times n}$ has finite order in $\operatorname{SL}(n,\mathbb Z)$, then $X = 0$. Given $p$ an odd prime, $X\in\mathbb{Z}^{n\times n}$ a matrix. 

If $I_{n}+pX\in \operatorname{SL}(n,\mathbb Z)$ has finite order, prove $X=0$.

My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$. 
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$, then $Y(hI+pZ)=0$. 
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? 
Thank you!
 A: Assume the claim is false. Let $m>1$ be minimal with the property that $X\ne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that 
 $$(I+pX)^a=I+\sum_{j=1}^a{a\choose j}p^{j}X^j =I+pX'$$
with $X':=\sum_{j=1}^a{a\choose j}p^{j-1}X^j\in\Bbb Z^{n\times n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$. 
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $X\ne 0$, let $r\ge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $Y\in\Bbb Z^{n\times n}$.
Then 
$$\tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{m\choose 2}p^{2r}Y^2+p^{3r}\cdot (\ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $m\choose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3\cdot(\ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
