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Is it possible to find a $n \times n$ matrix with integer entries of order $3n$?

This old answer by Gerry Myerson and the comments under it shed more light on the available orders of matrices in $GL_n(\Bbb{Z})$. In particular it follows that asymptotically the maximum order exceeds $3n$. However, the question remains, for which values of $n$ we can find matrices of order exactly $3n$?

Jyrki Lahtonen
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math maniac.
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  • Let $I$ be the $10\times10$ identity matrix, and let $\alpha=e^{2\pi i/15}$. Then $A=\alpha I$ has order 30. – Gerry Myerson Apr 26 '19 at 04:45
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    Alternatively, let $g$ be a primitive root modulo the prime $31$, let $I$ be the $10\times10$ identity matrix over the field of $31$ elements, then $A=gI$ has order $30$ over that field. – Gerry Myerson Apr 26 '19 at 04:47
  • Note that the question was edited to ask for integer entries after I posted the earlier comments. – Gerry Myerson Apr 26 '19 at 04:51
  • Yeah I have missed that point @Gerry Myerson. So sorry for that. – math maniac. Apr 26 '19 at 04:53
  • In my first comment, $\alpha$ should be $e^{\pi i/15}$. – Gerry Myerson Apr 26 '19 at 04:54
  • But that will not work here. Since $e^{\frac {\pi i} {15}}$ is not an integer. – math maniac. Apr 26 '19 at 04:56
  • It will work for the question as it was posed when I posted the comment. Also, it won't hurt anyone to learn something new. – Gerry Myerson Apr 26 '19 at 04:57
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    @mathmaniac. In the future, please refrain from making edits to your questions that - deliberately or otherwise - invalidate present answers. I am going to roll back your most recent edit; if you would like to ask the question you edited in, please ask it as a separate question. Also, please in the future include context (e.g. your own attempts, your own understanding, where the problem comes from) in your questions, since this question comes off as a "do my homework for me" type of thing without such context, and MSE does not take kindly to such behavior. – PrincessEev Apr 26 '19 at 05:33
  • Can I delete that question and ask the desired one @Eevee Trainer? – math maniac. Apr 26 '19 at 05:52
  • Since your question has already received an answer, no; better to keep it up in case anyone in the future needs it. Better to ask a new question – PrincessEev Apr 26 '19 at 06:43
  • The body and title of your question didn't match. So I edited your question. – little o Apr 26 '19 at 07:04
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    Any thoughts on the answers that have been posted? – Gerry Myerson Apr 30 '19 at 07:03
  • @EeveeTrainer In principle I agree with you. However, there are cases when the asker accidentally left out a key assumption. When that omission makes the question, calling it suspiciously trivial, then, in my opinion, it behooves the answerers to check by commenting. I would say that in the present case the omission made the question suspiciously trivial, but opinions may differ here. – Jyrki Lahtonen May 03 '19 at 04:15

2 Answers2

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Let $A=\pmatrix{0&1\cr1&0\cr}$, $B=\pmatrix{0&1&0\cr0&0&1\cr1&0&0\cr}$, $C$ the same thing but $5\times5$, then put those three matrices along the diagonal of a $10\times10$.

EDIT: The question has been changed again, to ask for an $n\times n$ integer matrix with order $3n$. This is possible for some $n$ such as $n=2$ and $n=10$ but not for others, e.g., there is no $3\times3$ integer matrix of order $9$.

I would encourage OP to do some experimenting to see whether there is any pattern in the $n$ for which one can/cannot find such a matrix.

Gerry Myerson
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  • Alternatively, find the minimal monic polynomial over the rationals for $e^{\pi i/15}$, then the companion matrix for this polynomial has integer entries and order $30$. It's $8\times8$, so you have to fill it out to a $10\times10$, say, by putting a couple of ones on the extended diagonal. – Gerry Myerson Apr 26 '19 at 04:56
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Construct a diagonal $10$ by $10$ matrix $M$ whose diagonal terms are from the $30^{th}$ roots of unity, then $M^{30}$ is the identity matrix. If at least one the diagonal terms has order of 30, then the matrix will have order of 30.

Mohammad Riazi-Kermani
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