# Prove that: if |det(A)|=1 and A^m=I for some m in bbbN^+, then we have A=I.

Let p be an odd prime number.
Let $A\in {M}_{n×n}\left(\mathbb{Z}\right)$ be a matrix satisfiying ${a}_{ij}\equiv {\delta }_{ij}\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}p\right)$
Prove that: if $|det\left(A\right)|=1$ and ${A}^{m}=I$ for some $m\in {\mathbb{N}}^{+}$, then we have $A=I$
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vidovitogv5
Hints.
We can assume that m is a prime (why?).
Now
${A}^{m}=\left(I+pX{\right)}^{m}=I+mpX+\left(\genfrac{}{}{0}{}{m}{2}\right){p}^{2}{X}^{2}+\dots =I$
If $m\ne p$ or m=p in any case $X=pY$ for some matrix Y.
Continuing also we obtain in the next step that $X={p}^{2}Z$ and so on.