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Let $S= \{A \in \text{Mat}_2(\mathbb{Q}) : A^6 = I$ and $A^n \ne I$ for any $0 < n < 6\}$.

I wish to describe the orbits of each of the element in $S$ with respect to conjugation by $GL_2(\mathbb{Q})$ on $\text{Mat}_2(\mathbb{Q})$.

I can't see how to start. Hints are much appreciated!

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3 Answers 3

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Every $A \in \operatorname{Mat}_2(\Bbb Q)$ has a conjugate element $PAP^{-1}$ (with $P \in GL_2(\Bbb Q)$) which is in its Frobenius normal form. Since $S$ is closed under conjugation, we can represent each orbit by enumerating the elements of $S$ which have this normal form.

Any element of $S$ will have a minimal polynomial which divides $x^6 - 1$. Over $\Bbb Q$, the polynomial $x^6 - 1$ factors into the product $$ p(x) = x^6 - 1 = (x-1)(x+1)(x^2 - x + 1)(x^2 + x + 1) $$ We eliminate any matrices whose minimal polynomial divides $x-1$, $x^2 - 1$, or $x^3 - 1$. The only monic polynomial over $\Bbb Q$ which divides $p$ but none of these smaller polynomials is $q(x) = x^2 - x + 1$. Thus, $q$ must be the minimal polynomial of our matrix in $S$.

Conclude that $S$ contains exactly one orbit, which belongs to the representative $$ A = \pmatrix{0&-1\\1&1} $$

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  • $\begingroup$ But when $k=3$, $A$ can have a $2$-by-$2$ block that looks like an element of $S$ and the remaining $1$-by-$1$ block is $1$ or $-1$. So, there are at least two conjugacy classes. (I think, for $k=3$, there are exactly $3$ conjugacy classes.) $\endgroup$ Aug 24, 2018 at 0:25
  • $\begingroup$ @Batominovski you're right; I was a bit careless in discussing the generalization. All we know is that the lcm of the relevant factors must be $x^6 - 1$. $\endgroup$ Aug 24, 2018 at 0:27
  • $\begingroup$ But it would be nice to find the number of conjugacy classes for an arbitrary dimension $k$, though. I'm too sleepy to do combinatorics now. Nice solution by the way. $\endgroup$ Aug 24, 2018 at 0:27
  • $\begingroup$ @Batominovski definitely an interesting problem now that you've pointed out this twist. If I find myself thinking about it later, I'll probably end up posting a question about the generalization. $\endgroup$ Aug 24, 2018 at 0:31
  • $\begingroup$ @Batominovski I ended up posting the question if you're interested $\endgroup$ Aug 24, 2018 at 18:14
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Let $A\in S$. From the condition that $A^6=I$ and $A^n\neq I$ for $0<n<6$, we conclude that the characteristic polynomial of $A$ must be $x^2-x+1$. We claim that $A$ is conjugate to $$X:=\begin{bmatrix}1&1\\ -1&0\end{bmatrix}\,.$$ That is, $S$ is a single orbit with representative $X$.

Suppose that $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\,,$$ where $a,b,c,d\in\mathbb{Q}$ are such that $a+d=1$ and $ad-bc=1$. Take $$V:=\begin{bmatrix}1&1\\c-d&-a+b\end{bmatrix}\,.$$ Then, $\det(V)=-a+b-c+d$.
First, we claim that $\det(V)\neq 0$. Suppose contrary that $\det(V)=0$. That is, $$a+c=b+d\,.$$ Since $d=1-a$, we have $$c=b-2a+1\,.$$ As $ad-bc=1$, we get $a(1-a)-b(b-2a+1)=1$, or $$(a-b)^2-(a-b)+1=0\,;$$ nonetheless, the equation $t^2-t+1=0$ has no solution $t\in\mathbb{Q}$. This is a contradiction, and the claim is proven.
Note that $$X\,V=\begin{bmatrix}1&1\\-1&0\end{bmatrix}\,\begin{bmatrix}1&1\\c-d&-a+b\end{bmatrix}=\begin{bmatrix}1+c-d&1-a+b\\-1&-1\end{bmatrix}$$ and $$V\,A=\begin{bmatrix}1&1\\c-d& -a+b\end{bmatrix}\,\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a+c&b+d\\-ad+bc&-ad+bc\end{bmatrix}\,.$$ Because $a+d=1$ and $ad-bc=1$, it can be seen immediately that $X\,V=V\,A$, whence $A$ is conjugate to $X$.

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Let $A$ be a matrix in $S$. Then its minimal polynomial divides the annihilator polynomial $x^6-1=(x^2 + x + 1)(x^2 - x + 1)(x + 1)(x - 1)$. If $A$ has a rational eigenvalue, than the other one is also rational, so both eigenvalues are among $\pm 1$, so $A^2=I$, contradiction. So $S$ has some non-rational eigenvalue. This must be a root of one of the factors $ (x^2 \pm x + 1)$, so the other eigenvalue is the Galois / complex conjugate. We exclude than the factor dividing $x^3-1$, because then $A^3$ would be $I$, contradiction. So the minimal polynomial of $A$ is $x^2-x+1$, thus we have a simpler characterization of $S$, $$ S=\left\{\ A=\begin{bmatrix}a& b\\ c& d\end{bmatrix}\ :\ \operatorname{Trace}(A)=a+d=1\ ,\ \det A= ad-bc=1\ \right\} \ . $$ Consider now two matrices $A,B$ from $S$ with entries $a,b,c,d$, and respectively $s,t,u,v$, $$ A = \begin{bmatrix}a& b\\ c& d\end{bmatrix} \ , \qquad B=\begin{bmatrix}s & t \\ u& v\end{bmatrix} \ . $$ We want (and show how) to conjugate them in each other. Of course, $bc\ne 0$, so a conjugation as follows $$ \begin{bmatrix}1& n\\ 0& 1\end{bmatrix} \begin{bmatrix}a& b\\ c& d\end{bmatrix} \begin{bmatrix}1& -n\\ 0& 1\end{bmatrix} = \begin{bmatrix}a+nc & *\\ *& *\end{bmatrix} \begin{bmatrix}1& -n\\ 0& 1\end{bmatrix} = \begin{bmatrix} a+nc & *\\ *& *\end{bmatrix} $$ can be found to reduce the situation to one with $a=s$. The trace condition implies $d=v$. Now $bc=tu$ (not zero) from the determinant condition, and a conjugation with a diagonal matrix (with diagonal $1$ and $t/b$) solves the problem.

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  • $\begingroup$ Here............ $\endgroup$
    – nonuser
    Aug 29, 2018 at 20:36

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