The group GL(2)

Let $F$ be any field. The set of $1$-dimensional subspaces of the $k + 1$-dimensional vector space $F^{k + 1}$ of column vectors having $k + 1$ rows is called projective $k$-space and denoted $\mathbb{P}^k (F)$. We can embed affine $k$-space $\mathbb{A}^k (F) = F^k$ into $\mathbb{P}^k (F)$ by mapping $(a_1, \cdots, a_k)$ to the one-dimensional subspace \[ F \left(\begin{array}{c} a_1\\ \vdots\\ a_k\\ 1 \end{array}\right) \; . \]

For example, if $F =\mathbb{R}$, the space $\mathbb{P}^k (\mathbb{R})$ is a compactification of affine space $\mathbb{R}^k$, whose introduction was motivated by perspective in art; thus the projective plane is the union of the affine plane with a "line at infinity.'' Projective geometry is more perfect than affine geometry; for example, in the affine plane, any two lines intersect in a single point, with one exception – parallel lines to not intersect. But in the projective plane, any two lines intersect in a single point, with no exceptions. (The line at infinity is where parallel lines intersect.)

Exercise 1.9.1: Show that the number of points in $\mathbb{P}^k (\mathbb{F}_p)$ is $p^k + p^{k - 1} + \ldots + 1$. This suggests a decomposition $\mathbb{P}^k (F) =\mathbb{A}^k (F) \cup \mathbb{A}^{k - 1} (F) \cup \cdots \cup \mathbb{A}^0 (F)$. Give a geometric interpretation.

The group $\operatorname{GL}(k + 1, F)$ acts on $F^{k + 1}$, and so it acts on $\mathbb{P}^k (F)$. However the center $Z$ consisting of scalar matrices \[ \left(\begin{array}{ccc} z & & \\ & \ddots & \\ & & z \end{array}\right) \] acts trivially, so the quotient group $\operatorname{GL}(k + 1,F) / Z = \operatorname{PGL}(k + 1,F)$ also acts.

The special case where $k = 1$ is already interesting. In this case, the projective line $\mathbb{P}^1 (F)$ consists of the affine line $F$ with a single point at infinity. An element $x \in F$ can be thought of as a point in the affine line $F$. It is identified with the one dimensional subspace \[ F \left(\begin{array}{c} x\\ 1 \end{array}\right) . \] To repeat: we embed $\mathbb{A}^1 (F) \longrightarrow \mathbb{P}^1 (F)$ by mapping $a \in \mathbb{A}^1 (F)$ to $F \left(\begin{array}{c} a\\ 1 \end{array}\right)$. These are all but one of the points in the projective plane. The last remaining point is \[ \infty = F \left(\begin{array}{c} 1\\ 0 \end{array}\right) . \] We will always identify $F =\mathbb{A}^1 (F)$ with its image in $\mathbb{P}^1 (F)$ and write $\mathbb{P}^1 (F) = F \cup \{\infty\}$.

Now let us consider the action of $\operatorname{GL}(2,F)$ on $\mathbb{P}^1 (F) = F \cup \{\infty\}$. We will see that it acts by linear fractional transformations. If \[ \gamma = \left(\begin{array}{cc} a & b\\ c & d \end{array}\right) \in \operatorname{GL}(2,F),\] it sends the vector $\binom{x}{1}$ to the vector \[ \left(\begin{array}{c} a x + b\\ c x + d \end{array}\right),\] so it sends the one dimensional space \[ F \left(\begin{array}{c} x\\ 1 \end{array}\right) \] to the one-dimensional space \[ F \left(\begin{array}{c} a x + b\\ c x + d \end{array}\right) = F \left(\begin{array}{c} \frac{a x + b}{c x + d}\\ 1 \end{array}\right) . \] Thus in the action on the projective line

\[ \gamma x = \frac{a x + b}{c x + d} . \]

(1.9.1)

For example, if $F =\mathbb{C}$ then $\mathbb{P}^1 (\mathbb{C}) =\mathbb{C} \cup \{\infty\}$ is topologically a sphere, the so-called Riemann sphere. The action of $\operatorname{PGL}(2,\mathbb{C})$ on the Riemann sphere is quite important in complex analysis.

But for us, $F =\mathbb{F}_q$ will be a finite field. We note that we have constructed $\mathbb{F}_p$ when $p$ is a prime as the quotient ring $\mathbb{Z}/ p\mathbb{Z}$. But Galois proved that if $q = p^n$ there is a field $\mathbb{F}_q$. It is of course not the finite ring $\mathbb{Z}/ p^n \mathbb{Z}$, which is not a field if $n > 1$. We will not discuss the construction of the general finite fields, but the field $\mathbb{F}_4$ will appear in one of the exercises.

The order of $\operatorname{GL}(2,\mathbb{F}_q)$ is $(q^2 - 1) (q^2 - q)$, and the order of both groups $\operatorname{PGL}(2,\mathbb{F}_q)$ is the same, $q (q^2 - 1)$. The order of $\operatorname{PSL}_2 (\mathbb{F}_q)$ is $\frac{1}{2} q (q^2 - 1)$ if $q$ is odd, but if $q$ is even, then $\operatorname{PSL}_2 (\mathbb{F}_q)$ and $\operatorname{SL}_2 (\mathbb{F}_q)$ are really the same group, since the center of $\operatorname{SL}_2 (\mathbb{F}_q)$ is trivial. (See Exercises 1.5.6, 1.5.7 and 1.5.8.) Here is a table of some values.

The orders of some small groups, including simple groups of orders $60, 168, 503$ and 360.

For comparison, we have listed the values of $S_n$ and $A_n$ for some small values of $n$. We will discuss some "accidental'' isomorphisms between projective groups and alternating groups. Such isomorphisms only happen for small $n$ – after a few such isomorphisms at the beginning, the infinite series of simple groups produce nonisomorphic groups. However, these accidental isomorphisms are examples of a very general phenomenon, which is that generally every group has very different concrete realizations, and this may be regarded as a "theme'' in representation theory.

The group $\operatorname{PGL}(2,\mathbb{F}_q)$ is embedded in $S_{q + 1}$ by virtue of its faithful action on $\mathbb{P}^1 (\mathbb{F}_q)$. Thus $\operatorname{PGL}(2,\mathbb{F}_2)$ is embedded in $S_3$, and since both groups have the same order 6, it is obvious that $\operatorname{PGL}(2,\mathbb{F}_2) \cong S_3$. Similarly $\operatorname{PGL}(2,\mathbb{F}_3)$ is embedded in $S_4$, and both groups have the same order $24$. So $\operatorname{PGL}(2,\mathbb{F}_3) \cong S_4$.

We have not proved Galois' theorem that there exists a field of order $q$ for any prime power $q = p^n$, but when $q = 4$ it is simple enough just to write down the additive and multiplicative laws. There are four elements, $\{0, 1, \alpha, \beta\}$ and the addition and multiplication tables are as follows. \[ \begin{array}{|l|l|l|l|l|} \hline + & 0 & 1 & \alpha & \beta\\ \hline 0 & 0 & 1 & \alpha & \beta\\ \hline 1 & 1 & 0 & \beta & \alpha\\ \hline \alpha & \alpha & \beta & 0 & 1\\ \hline \beta & \beta & \alpha & 1 & 0\\ \hline \end{array} \hspace{4em} \begin{array}{|l|l|l|l|l|} \hline \times & 0 & 1 & \alpha & \beta\\ \hline 0 & 0 & 0 & 0 & 0\\ \hline 1 & 0 & 1 & \alpha & \beta\\ \hline \alpha & 0 & \alpha & \beta & 1\\ \hline \beta & 0 & \beta & 1 & \alpha\\ \hline \end{array} \] This is all the proof that we will offer that this field exists.

The group $\operatorname{PGL}(2,\mathbb{F}_4)$ is embedded in $S_5$, and has order 60, so we might surmise that it equals $A_5$, but how to prove it? In this case, there is an easy argument that avoids calculation. If $\operatorname{PGL}(2,\mathbb{F}_4) \subset S_5$ were distinct from $A_5$, then its intersection with $A_5$ would have order 30. To see this, observe that the sign character $\varepsilon : S_5 \longrightarrow \{\pm 1\}$ would be nontrivial when restricted to $\operatorname{PGL}(2,\mathbb{F}_4)$, so its kernel $A_5 \cap \operatorname{PGL}(2,\mathbb{F}_4)$ would be a group of order 30, that is, a subgroup of index 2 in $A_5$. This would be a normal subgroup of $A_5$ (Exercise 1.7.4) contradicting Proposition 1.7.2.

The next surmise that one might make from comparing the orders of the groups in Table 1.9.2 is that perhaps $\operatorname{PSL}_2 (\mathbb{F}_5) \cong A_5$ and $\operatorname{PGL}(2,\mathbb{F}_5) \cong S_5$. Proving this is harder, because the natural permutation action of $\operatorname{PGL}(2,\mathbb{F}_5)$ is on a set with $6$ elements, while the natural permutation action of $S_5$ is on a set with 5 elements.

One seeks therefore a second permutation action of $S_5$ on something else. Is there a natural transitive permutation of $S_5$ on a set with six elements? A good answer is to make use of the $5$-Sylow subgroups. There are six of these.

We will label the six 5-Sylows by elements of $\mathbb{P}^1 (\mathbb{F}_5)$. It will be convenient to think of $S_5$ as $\operatorname{Bij} (\{a, b, c, d, e\})$ so as to avoid confusion since $0, 1, 2, 3, 4$ appear as elements of $\mathbb{P}^1 (\mathbb{F}_5)$. The Sylows, labeled thus, are: \begin{eqnarray*} \left\langle (a b c d e) \right\rangle & = & \infty\\ \left\langle (a b c e d) \right\rangle & = & 0\\ \left\langle (a b d e c) \right\rangle & = & 1\\ \left\langle (a b e d c) \right\rangle & = & 2\\ \left\langle (a b e c d) \right\rangle & = & 3\\ \left\langle (a b d c e) \right\rangle & = & 4 \end{eqnarray*} This labeling was chosen so as to make the following proof work.

Exercise 1.9.2: Prove that $(a b)$ and $(a b c d e)$ generate $S_5$.

Exercise 1.9.3: Compute the effect of $(a b)$ and $(a b c d e)$ on the six 5-Sylow subgroups of $S_5$. With the above labeling, find two linear fractional transformations that have the same effect, and explain how this shows that $\operatorname{PGL}(2,\mathbb{F}_5) \cong S_5$.

Now let us discuss the conjugacy classes of $GL(2,\mathbb{F}_q)$. In general, the conjugacy classes of $GL(n,\mathbb{F}_q)$ have a simpler description than the conjugacy classes of $SL(n,\mathbb{F}_q)$, just as the conjugacy classes of $S_n$ have a simpler description than the conjugacy classes of $A_n$. For $GL(2)$, there are really just four types of conjugacy classes.

One of the four conjugacy classes requires a bit of preparation. Let us recall some further facts about finite fields. We have mentioned that Galois proved that if $q$ is a prime power then there is a finite field of order

We have mentioned that if $q$ is a prime power then Galois proved that there exists a finite field of order $q$, to be denoted $\mathbb{F}_q$. By the same token there is a field $\mathbb{F}_{q^n}$ of order $q^n$. It contains $\mathbb{F}_q$. The Galois group $\operatorname{Gal} (\mathbb{F}_{q^n} /\mathbb{F}_q)$ is by definition the group of field automorphisms of $\mathbb{F}_{q^n}$ that are trivial on $\mathbb{F}_q$. It is a cyclic group of order $n$, generated by the Frobenius automorphism $\phi : \mathbb{F}_{q^n} \longrightarrow \mathbb{F}_{q^n}$, that is defined by $\phi (x) = x^q$. For example, $\operatorname{Gal} (\mathbb{F}_4 /\mathbb{F}_2)$ is generated by $\phi (x) = x^3$, and with notation as above, this is the map whose values are given by the following table. \[ \begin{array}{|l|l|l|l|l|} \hline x & 0 & 1 & \alpha & \beta\\ \hline \phi (x) & 0 & 1 & \beta & \alpha\\ \hline \end{array} \]

Let $\alpha \in \mathbb{F}_{q^n}^{\times}$. Let $\lambda_{\alpha} : \mathbb{F}_{q^n} \longrightarrow \mathbb{F}_{q^n}$ be the map $\lambda_{\alpha} (x) = \alpha x$. Now $\mathbb{F}_{q^n}$ is an $n$-dimensional vector space over $\mathbb{F}_q$, and $\lambda_{\alpha}$ is an invertible linear transformation of this vector space. So if we choose an $\mathbb{F}_q$-vector space basis of $\mathbb{F}_{q^n}$, we may identify \[ \operatorname{End} (\mathbb{F}_{q^n}) = \operatorname{Mat}_n (\mathbb{F}_q), \] and the element $\lambda_{\alpha}$ becomes an element of $\operatorname{GL} (n, \mathbb{F}_q)$.

Theorem 1.9.1: The conjugacy classes of $\operatorname{GL} (2, \mathbb{F}_q)$ are as follows.

(iv) If $\alpha \in \mathbb{F}_{q^2} -\mathbb{F}_q$ then the conjugacy class of $\lambda_{\alpha}$ has $q^2 - q$ elements. The elements $\lambda_{\alpha}$ and $\lambda_{\phi (\alpha)}$ are conjugate. There are $\frac{1}{2} (q^2 - q)$ conjugacy classes of this type.

Proof. (Click to Expand/Collapse)