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Problem of the Week #49 - May 6th, 2013

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Chris L T521

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Jan 26, 2012
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Here's this week's problem.

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Problem: Let $\mathbb{F}_q$ be a field with $q$ elements. How many conjugacy classes are there in $\mathrm{GL}_2(F_q)$? Use rational canonical form, considering the two cases of a cyclic $\mathbb{F}_q$-module and a non-cyclic $\mathbb{F}_q$-module separately.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
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Jan 26, 2012
995
No one answered this week's problem. You can find my solution below.

Solution: Let $\phi \in \mathrm{GL}_2(\mathbb{F}_q)$.

If $\mathbb{F}_q^2$ is a cyclic $\mathbb{F}_q[x]$- module associated with $\phi$, choose a basis for which $\phi$ is represented by
\[\begin{pmatrix}0 & 1 \\-a_0 & -a_1\end{pmatrix}\]
where $P^{\phi}_{min} = X^2 + a_1 X + a_0$ is the minimal polynomial for $\phi$.


Since $\phi \in \mathrm{GL}_2(\mathbb{F}_q)$, then $a_0 \neq 0$. Hence for $a_0$, there are $q-1$ possibilities. And $a_1$ can be arbitrary, then there are $q$ possibilities. Therefore, the total number of different conjugacy classes is $q(q-1)$.


If $\mathbb{F}_q^2$ is not a cyclic $\mathbb{F}_q[x]$- module associated with $\phi$, choose a basis for which $\phi$ is represented by
\[\begin{pmatrix}\alpha & 0 \\ 0 & \beta\end{pmatrix}\]


If $\alpha = \beta$, then $\mathbb{F}_q^2$ is a cyclic $\mathbb{F}_q[x]$- module associated with $\phi$ and it contradicts to the presumption here. Hence $\alpha \neq \beta$ and since it is in $\mathrm{GL}_2(\mathbb{F}_q)$, then there are $q-1$ possibilities for $\alpha$ and $\beta$, respectively. Therefore, the total number of different conjugacy classes is $(q-1)^2$.


Now in summary, the total number of different conjugacy classes are $q(q-1) + (q-1)^2 = q^2 - 1.\quad\clubsuit$
 
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