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I would want to make an example of a matrix $M \in GL_3(\mathbb{Z}_7)$ such that $\langle M; +, \cdot \rangle \simeq GF(7^3)$ and the multiplicative order of $M$ is $3$.

Any hints how to do that with a computer algebra system would be appreciated.

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    Are you sure that you copied the problem correctly? If $M$ has multiplicative order $3$, and $M$ belongs to a field of characteristic seven, then $M$ has to be in the prime field, because $x^3-1=(x-1)(x-2)(x-4)$ modulo 7. Therefore $M$ will not generate a cubic extension. Please check/comment! BTW I find the notation $\langle M;+,\cdot\rangle$ a little bit non-standard. I assume that it means the structure generated by $M$ that is closed under matrix addition and multiplication. Is this correct?2011-10-27
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    Thank you for the answer! I am sorry for the late reply. I have been ill. I indeed meant by ⟨M;+,⋅⟩ the structure generated by $M$ that is closed under matrix addition and multiplication. Let us denote it by $S$. Actually I would want to make an example of a matrix $M \in GL_3(\mathbb{Z}_p)$, where $p$ is a prime number, such that $M$ has prime order $q$, $q \neq p$, and $S \simeq GF(p^3)$. If there are not any such matrices for some $p$ and $q$, then it is ok as well. I do not understand at the moment how exactly you implied from that equation that $M$ must be in the prime field.2011-11-01
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    Ok. For the field $GF(p^3)$ to have an element of multiplicative order $q$ it is necessary and sufficient that $q\mid p^3-1$. This is because the multiplicative group of $GF(p^3)$ is cyclic of order $p^3-1$. What went wrong in your example case was that with $q=3$, $p=7$ we already have $q\mid p-1$. Therefore the elements of order $q$ all belong to the smaller field $GF(p)$. So we can say that such a matrix $M$ exists, if and only if $q\mid p^3-1$, but $q\nmid p-1$.2011-11-01
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    Finding such a matrix $M$ is more difficult. If your CAS can factor $x^q-1$ over $GF(p)$, then I would pick a cubic factor and use a companion matrix of that.2011-11-01
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    My argument leading to the conclusion that $M$ has to be in the prime field only works, if we already know that the algebra genereated by $M$ is a field. The example from Jack's answer shows that it is possible to find a matrix $M$ of order 3 that generates an algebra of the correct size. That matrix is not in the prime field (=not a scalar matrix), but the algebra that it generates isn't a field.2011-11-02
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    I have to study some notation that I am not so used to in the first paragraph of the answer you posted to understand some blanks I have. Factoring $x^q-1$ over $GF(p)$ in GAP should be possible. I will inform about the progress later. I appreciate the hints.2011-11-02
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    I seem to have left out that what always holds is $$\langle M,+,\cdot\rangle\simeq GF(p)[x]/\langle \chi_M(x)\rangle,$$ where $\chi_M(x)$ is the characteristic polynomial of $M$.2011-11-03

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