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Chapter 6: Problem 12

Let \(F\) be a finite field with \(q\) elements. Show that the order of \(\mathrm{GL}_{2}(F)\) is \(q\left(q^{2}-1\right)(q-1)\)

Short Answer

Expert verified
The order of the general linear group \(\mathrm{GL}_{2}(F)\) over a finite field \(F\) with \(q\) elements can be determined by counting the number of \(2\times 2\) invertible matrices. First, we count the possibilities for the first column, which can be any non-zero vector in \(F^2\), resulting in \(q^2 - 1\) possibilities. Second, we count the alternatives for the second column, which must be linearly independent from the first column; hence, there are \(q^2 - q\) possibilities. Multiplying these amounts together, we obtain \((q^2 - 1)(q^2 - q)\) total possibilities. Factoring out a common factor of \(q\) results in the expression \(q(q^2 - 1)(q - 1)\). Hence, the order of \(\mathrm{GL}_{2}(F)\) is \(q\left(q^{2}-1\right)(q-1)\).

Step by step solution

01

Counting the number of possibilities for the first column of the matrix

Since we want an invertible matrix, the first column of the matrix can be any non-zero vector in \(F^2\). There are \(q^2\) possible vectors in total, and since the zero vector is not allowed, we have \(q^2 - 1\) possibilities for the first column.
02

Counting the number of possibilities for the second column of the matrix

Now that we have chosen the first column of the matrix, the second column must be linearly independent from the first column to ensure that the matrix is invertible. The second column can be any vector that is not a scalar multiple of the first column. Since there are \(q\) possible scalar multiples of the first column (including the zero vector), there are \(q^2 - q\) possibilities for the second column.
03

Calculating the total number of possibilities for both columns

To find the total number of possible matrices in \(\mathrm{GL}_{2}(F)\), we multiply the number of possibilities for each column. So, the total number of possibilities is \((q^2 - 1)(q^2 - q)\).
04

Factor out the common factor

Observe that both terms in the product have a common factor of \(q\). We can factor out this common factor to get \[(q^2 - 1)(q^2 - q) = q(q^2 - 1)(q - 1).\]
05

Conclusion

We have shown that there are \(q\left(q^{2}-1\right)(q-1)\) possible \(2\times 2\) invertible matrices with entries from the finite field \(F\). Therefore, the order of \(\mathrm{GL}_{2}(F)\) is \(q\left(q^{2}-1\right)(q-1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Finite Field
Imagine a mathematical universe of numbers where the quantity is limited, and yet within this limited set, you can still perform addition, subtraction, multiplication, and even division (except by zero), and all results stay within this set. This is essentially what a finite field is; a finite set of elements that form a field. Specifically, in the exercise, we're dealing with a finite field with exactly q elements.

A finite field, also known as a Galois field, is paramount in various applications, especially in cryptography and coding theory, because of its circular nature. In a finite field, every operation you do brings you back to another element within the same field, much like a game of musical chairs with precisely q seats, and every participant is guaranteed a seat regardless of the music's tempo. It provides a controlled, predictable environment for complex calculations where every element has a unique inverse regarding addition and multiplication.
Invertible Matrices
Let's dive into the world of flipping matrices on their heads, also known as finding the inverse of matrices. Invertible matrices are square matrices that have a unique counterpart such that when multiplicated together, result in the identity matrix. This is vital because in matrix algebra, multiplication does not generally have the luxury of division. So, the inverse is our knight in shining armor.

In the context of our exercise, we're exploring 2x2 matrices over a finite field. To qualify as an invertible matrix, this special square must pass a major test: its determinant cannot be zero. Otherwise, it's like trying to divide by zero 鈥� a big no-no. Now, when we mix invertible matrices with finite fields, it's a match made in heaven. It ensures we are dealing with a finite set of possibilities, making the enumeration of these matrices a game with clear rules.
Linear Independence
Just as individuals crave their uniqueness, vectors in a matrix seek their own independence, known as linear independence. But what does it mean for vectors to be linearly independent? Imagine two vectors that don't fall in line with each other, neither one being a simple rescale of the other. They refuse to follow the same path and together can point to any spot in their vector space.

In the problem we're tackling, the matrices are 2x2, meaning they have two columns, and we want these two vectors (columns) to be linearly independent. This is crucial because only with this rebellion against duplication can the matrix claim the noble title of 'invertible'. If these vectors were linearly dependent (like the conga line), the matrix would become singular, which in matrix speak, is the doom of no return (non-invertible). So, when we choose these vectors with the care of picking teammates for a relay race, we ensure each brings something different to the table, covering every possible point in the race against dependency.

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Most popular questions from this chapter

Let \(\mathrm{H}\) be the upper half plane, that is the set of all complex numbers $$z=x+i y$$ with \(y>0\). Let $$\alpha=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \in G$$ Define $$\alpha(z)=\frac{a z+b}{c z+d}$$ Prove by explicit computation that \(\alpha(z) \in \mathbf{H}\) and that: (a) If \(\alpha, \beta \in G\) then \(\alpha(\beta(z))=(\alpha \beta)(z)\). (b) If \(\alpha=\pm I\) then \(\alpha(z)=z\). In other words, we have defined an operation of \(\mathrm{SL}_{2}(\mathbf{R})\) on \(\mathbf{H}\), according to the definition of Chapter II, \(\$ 8\).

Let \(\mathrm{SL}_{n}(R)\) be the subset of \(\mathrm{G} \mathrm{L}_{-n}(R)\) consisting of those matrices with determinant 1. Show that \(\mathrm{SL}_{n}(R)\) is a subgroup.

(a) Let \(G=G_{0}\) be the group of \(3 \times 3\) upper triangular matrices in a field \(K\), consisting of all invertible triangular matrices $$ T=\left(\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33} \end{array}\right) $$ so \(a_{11} a_{22} a_{33} \neq 0 .\) Let \(G_{t}\) be the set of matrices $$ \left(\begin{array}{ccc} 1 & a_{12} & a_{13} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \end{array}\right) $$ Show that \(G_{1}\) is a subgroup of \(G\), and that it is the kernel of the homomorphism which to each triangular matrix \(T\) associates the diagonal matrix consisting of the diagonal elements of \(T\). (b) Let \(G_{2}\) be the set of matrices $$ \left(\begin{array}{lll} 1 & 0 & c \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) $$ Show that \(G_{2}\) is a subgroup of \(G_{1}\). (c) Generalize the above to the case of \(n \times n\) matrices. (d) Show that the map $$ \left(\begin{array}{ccc} 1 & a_{12} & a_{13} \\ 0 & 1 & a_{23} \\ 0 & 0 & 1 \end{array}\right) \mapsto\left(a_{12}, a_{23}\right) $$ is a homomorphism of the group \(G_{1}\) onto the direct product $$ K \times K=K^{2} $$ What is the kernel? (e) Show that the group \(G_{2}\) is isomorphic to \(K\).

Let \(G=\mathrm{GL}_{2}(F)\) where \(F=\mathrm{Z} / 3 \mathrm{Z}\), and let \(V=F \times F\) be the vector space of pairs of elements of \(F\), having dimension 2 over \(F\). (a) Show that \(G\) operates as a permutation group of the subspaces of \(V\) of dimension 1. How many such subspaces are there? (b) From (a), establish an isomorphism \(G / \pm 1 \approx S_{4}\) (where \(S_{4}\) is the symmetric group on 4 clements). (c) Establish an isomorphism \(\mathrm{SL}_{2}(\mathbf{Z} / 3 \mathbf{Z}) / \pm 1 \approx A_{4} .\) where \(A_{4}\) is the alternating subgroup of \(S_{4}\).

Let \(F\) be a finite field with \(q\) elements. What is the order of the group of diagonal matrices: (a) \(\left(\begin{array}{ll}a & 0 \\ 0 & d\end{array}\right)\) with \(a, d \in F, a d \neq 0\) (b) \(\left(\begin{array}{cccc}a_{1} & 0 & \ldots & 0 \\ 0 & a_{2} & \ldots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \ldots & a_{n}\end{array}\right)\) with \(a_{1} \ldots . a_{n} \in F, a_{i} \neq 0\) for all \(i\)
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