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

Zack Chase 2022-09-25 Answered
Let p be an odd prime number.
Let A M n × n ( Z ) be a matrix satisfiying a i j δ i j ( mod p )
Prove that: if | det ( A ) | = 1 and A m = I for some m N + , then we have A = I
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Answers (1)

vidovitogv5
Answered 2022-09-26 Author has 10 answers
Hints.
We can assume that m is a prime (why?).
Now
A m = ( I + p X ) m = I + m p X + ( m 2 ) p 2 X 2 + = I
If m p or m=p in any case X = p Y for some matrix Y.
Continuing also we obtain in the next step that X = p 2 Z and so on.
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