How many conjugacy classes in $\mathrm{GL}_2(F_q)$?

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In summary, the number of conjugacy classes in $\mathrm{GL}_2(F_q)$ can be determined by finding the number of distinct sizes of Jordan blocks in the corresponding Jordan canonical form for each matrix in the group. This can be achieved by calculating the size of the minimal polynomial for each matrix, which will give the number of Jordan blocks with distinct eigenvalues. There is a formula for calculating the number of conjugacy classes in $\mathrm{GL}_2(F_q)$, which is $\frac{1}{q^2-1} \sum_{d|n} \mu(d)q^{2n/d}$, where $q$ is the order of the finite field $F_q$ and $n$ is the
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Chris L T521
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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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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 $q(q-1) + (q-1)^2 = q^2 - 1.\quad\clubsuit$
 

FAQ: How many conjugacy classes in $\mathrm{GL}_2(F_q)$?

How do you determine the number of conjugacy classes in $\mathrm{GL}_2(F_q)$?

The number of conjugacy classes in $\mathrm{GL}_2(F_q)$ can be determined by finding the number of distinct sizes of Jordan blocks in the corresponding Jordan canonical form for each matrix in the group. This can be achieved by calculating the size of the minimal polynomial for each matrix, which will give the number of Jordan blocks with distinct eigenvalues.

Is there a formula for calculating the number of conjugacy classes in $\mathrm{GL}_2(F_q)$?

Yes, there is a formula for calculating the number of conjugacy classes in $\mathrm{GL}_2(F_q)$. It is given by the formula:

$$\frac{1}{q^2-1} \sum_{d|n} \mu(d)q^{2n/d}$$

where $q$ is the order of the finite field $F_q$ and $n$ is the degree of the matrix in $\mathrm{GL}_2(F_q)$.

How does the number of conjugacy classes in $\mathrm{GL}_2(F_q)$ relate to the order of the group?

The number of conjugacy classes in $\mathrm{GL}_2(F_q)$ is related to the order of the group through the class equation. The class equation states that the order of the group is equal to the sum of the orders of its conjugacy classes. Therefore, by knowing the number of conjugacy classes, we can determine the order of the group.

Can the number of conjugacy classes in $\mathrm{GL}_2(F_q)$ vary for different finite fields $F_q$?

Yes, the number of conjugacy classes in $\mathrm{GL}_2(F_q)$ can vary for different finite fields $F_q$. This is because the number of distinct Jordan block sizes and therefore, the number of conjugacy classes, depends on the order of the finite field $q$.

How is the number of conjugacy classes in $\mathrm{GL}_2(F_q)$ related to the number of elements in the field $F_q$?

The number of conjugacy classes in $\mathrm{GL}_2(F_q)$ is related to the number of elements in the field $F_q$ through the Jordan-Hölder theorem. This theorem states that for any finite group, the number of conjugacy classes is equal to the number of irreducible representations of the group, which is equivalent to the number of elements in the field $F_q$.

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