Is $A^4=1$ for a $3\times 3$ matrix over $\mathbb{Q}$ if $A^8=1$?

[Chris L T521]

Here's this week's problem (and the first Graduate POTW of 2013!).

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Problem: Prove that if $A$ is a $3\times 3$ matrix over $\mathbb{Q}$ such that $A^8=1$, then in fact $A^4=1$.

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[Chris L T521]

This week's problem was correctly answered by Deveno and Opalg. You can find Opalg's solution below:

Spoiler

First, notice that if $A$ is considered as an element of $M_3(\mathbb{C})$ it is diagonalisable because it has finite order. (If its Jordan form had any off-diagonal elements, it could not have finite order.) Its eigenvalues are 8th roots of unity, and we have to show that in fact they are 4th roots of unity.

The irreducible factorisation of $A^8-I=0$ over $\mathbb{Q}$ is $(A-I)(A+I)(A^2+I)(A^4+I) = 0$. The polynomial $(x-1)(x+1)(x^2+1)(x^4+1)$ must therefore be a multiple of the minimal polynomial of $A$ The characteristic polynomial $p(x)$ of $A$ must be a product formed using those same irreducible factors, namely $x\pm1$, $x^2+1$ and $x^4+1.$ It has degree 3, so it cannot involve the factor $x^4+1.$ So it must be formed from copies of the other factors, whose roots are all 4th roots of unity.