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If A is a rational matrix $n$ x $n$ matrix with $n \leq 3$ and has finite order in $GL_n(\mathbb{Q})$, then it has order 1,2,3,4, or 6. I am trying to prove this statement using field theory. I have observed that 1,2,3,4 and 6 are the only numbers with totient at most 3. If $A$ has order $k$, then it's eigenvalues are $k$th roots of unity. However, the fact that they are not necessarily primitive $k$th roots of unity is giving me trouble. Any ideas for how to prove this claim?

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Hint: If an eigenvalue is a primitive $j$'th root of unity with $j=3,4$ or $6$, it and its conjugate are two eigenvalues, and the other eigenvalue must be $1$ or $-1$.

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