Lecture 6: Friday, August 30, 2013

Groups were first studied as objects acting on sets. For example, we can consider the group of rotations of a regular polygon. Eventually, we wish to consider a specific type of group acting on the collection $K(z) = \left \{ \left. f(z) = \dfrac {a_n \, z^n + \cdots + a_1 \, z + a_0}{b_m \, z^m + \cdots + b_1 \, z + b_0} \ \right| \ a_i, \, b_j \in K \right \}$ of rational functions $f(z)$ over a field $K$ . Today, we review various definitions and properties by focusing exclusively on dihedral, symmetric, alternating, and matrix groups.

Definitions of Groups

Let $G$ be a set, and assume that there is some composition “ $\circ$ ” on $G$ ; that is, given $a, \, b \in G$ there is a unique element $a \circ b \in G$ . We say that $G$ is a group under $\circ$ if the following four properties hold:

Closure: Given $a, \, b \in G$ the composition $a \circ b \in G$ .

Associativity: Given $a, \, b, \, c \in G$ we have $a \circ (b \circ c) = (a \circ b) \circ c$ . (This is to make sure the expression $a \circ b \circ c$ is unambiguous.)

Identity: There is a unique element $e \in G$ such that for all $a \in G$ we have $a \circ e = e \circ a = a$ . (We call $e$ the identity element under $\circ$ .)

Inverses: Given $a \in G$ there is a unique element $b \in G$ such that $a \circ b = b \circ a = e$ . (We call $b$ the inverse of $a$ , and sometimes write $b = a^{-1}$ or $b = -a$ depending on the operation $\circ$ .)

We say that $G$ is an abelian group if (Closure) — (Inverses) hold, and additionally

Commutativity: Given $a, \, b \in G$ we have $a \circ b = b \circ a$ .

We say $G$ is finitely generated if there exists a finite set $S = \{ s_1, s_2, \dots, s_k \} \subseteq G$ such that, for any $a \in G$ , there exist integers $1 \leq i_j \leq k$ such that $a = s_{i_1} \circ s_{i_2} \circ \cdots \circ s_{i_m}$ . The set $S$ is called a generating set for the group $G$ ; this set is not unique. We often write $G = \langle S \, | \, R \rangle$ , where $R$ is the collection of ways to express $s_{i_1} \circ s_{i_2} \circ \cdots \circ s_{i_m} = e$ . This is a presentation for the group $G$ ; it is also not unique.

Dihedral Groups

We give a fundamental example of a group which is not abelian.

Theorem.
For each positive integer $n$ , let the dihedral group $D_{n}$ denote the set of symmetries of the regular $n$ -gon $V = \left \{ \left. P_k = \left( \cos \, \dfrac {2 \pi k}n, \ \sin \, \dfrac {2 \pi k}n \right) \in \mathbb A^2(\mathbb R) \ \right| \ k \in \mathbb Z \right \}$ .

The group $D_{n}$ has $2 \, n$ elements.

It can be generated by just two elements, namely the rotation $r$ of $(2\pi/n)$ radians counterclockwise and a reflection $s$ about the line $y = m \, x$ of slope $m = \tan (2 \pi/n)$ .

In particular, a presentation for the dihedral group is $D_n = \left \langle r, \, s \, \bigl| \, s^2 = r^n = (s \, r)^2 = 1 \right \rangle$ .

Proof: First, let’s write down some obvious symmetries of the $n$ -gon. We have a rotation $r: P_k \mapsto P_{k+1}$ as well as a reflection $s: P_k \mapsto P_{2-k}$ about the line $y = m \, x$ . We can express both in terms of matrices: $r = \left [ \begin{matrix} \cos \, \dfrac {2 \pi}n & -\sin \, \dfrac {2 \pi}n \\ \\ \sin \, \dfrac {2 \pi}n & \cos \, \dfrac {2 \pi}n \end{matrix} \right ]$ and $s = \left [ \begin{matrix} \dfrac {1-m^2}{1+m^2} & \dfrac {2 m}{1+m^2} \\ \\ \dfrac {2 m}{1+m^2} & -\dfrac {1-m^2}{1+m^2} \end{matrix} \right ] = \left [ \begin{matrix} \cos \, \dfrac {4 \pi}n & \sin \, \dfrac {4 \pi}n \\ \\ \sin \, \dfrac {4 \pi}n & -\cos \, \dfrac {4 \pi}n \end{matrix} \right ]$ . It is easy to check that $s^2 = r^n = (s \, r)^2 = 1$ .

Now we can count the number of symmetries. Any symmetry $g \in D_{n}$ must be a rigid rotation of the regular $n$ -gon, so it is uniquely determined by where it sends the two vertices $P_1$ and $P_2$ . Say that $g: P_1 \mapsto P_k$ . Since $P_2$ is adjacent to $P_1$ , it must go to $P_{k \pm 1}$ . The symmetry $r^{k-1}: P_1 \mapsto P_k$ , while $r^{k-1}: P_2 \mapsto P_{k+1}$ and $r^{k-1} s: P_2 \mapsto P_{k-1}$ . Hence either $g = r^{k-1}$ or $g = r^{k-1} \, s$ . There are $2 \, n$ such choices. $\square$

Morphisms between Groups

Say that $G$ and $H$ are groups under the composition laws $\circ$ and $\diamond$ , respectively. We say that a well-defined map $\phi: G \to H$ is a group homomorphism if it preserves the group operations, that is, $\phi(x \circ y) = \phi(x) \diamond \phi(y)$ . The set of group homomorphisms from $G$ to $H$ is denoted by $\text{Hom}(G, H)$ . We say that a group homomorphism is an isomorphism if it is a bijection. If $\phi:G \to H$ is indeed an isomorphism, we say that $G$ and $H$ are isomorphic and write $G \simeq H$ . Similarly, a group homomorphism $\gamma: G \to G$ is called an endomorphism; a bijective endomorphism is called an automorphism. We denote these sets by $\text{End}(G) = \text{Hom}(G,G)$ and $\text{Aut}(G)$ , respectively.

Proposition.
Let $G$ , $H$ , and $K$ be groups under the composition laws $\circ$ , $\diamond$ , and $\star$ , respectively. Fix $\phi \in \text{Hom}(G, H)$ and $\psi \in \text{Hom}(H, K)$ .

The composition $\psi \circ \phi \in \text{Hom}(G, K)$ . In particular, $\text{Aut}(G)$ is a group under composition.

Say that $e$ is the identity of $G$ and $\epsilon$ is the identity of $H$ . Denote the kernel and the image, respectively, of the homomorphism $\phi$ as the sets $\text{ker} \, \phi = \left \{ a \in G \, \bigl| \, \phi(a) = \epsilon \right \}$ and $\phi(G) = \left \{ \alpha \in H \, \bigl| \, \alpha = \phi(a) \ \text{for some} \ a \in G \right \}$ . Then $\text{ker} \, \phi$ is a subgroup of $G$ while $\phi(G)$ is a subgroup of $H$ .

Proof: Say that $\phi: G \to H$ and $\psi: H \to K$ . Clearly $\psi \circ \phi: G \to K$ is a well-defined map. For $a, \, b \in G$ we have
$\begin{aligned} \bigl( \psi \circ \phi \bigl)(a \circ b) & = \psi \bigl( \phi(a \circ b) \bigr) \\[5pt] & = \psi \bigl( \phi(a) \diamond \phi(b) \bigr) \\[5pt] & = \psi \bigl( \varphi(a) \bigr) \star \psi \bigl( \varphi(b) \bigr) \\[5pt] & = \bigl( \psi \circ \varphi \bigr)(a) \star \bigl( \psi \circ \varphi \bigr)(b). \end{aligned}$
Hence $\psi \circ \varphi$ is also a group homomorphism.

To show $\text{Aut}(G)$ is a group under composition, we must show that four axioms are satisfied. To see why (Closure) holds, note that the composition $\gamma_1 \circ \gamma_2 \in \text{End}(G)$ for any $\gamma_1, \, \gamma_2 \in \text{Aut}(G)$ . The composition of two bijections is also a bijection because $\left( \gamma_1 \circ \gamma_2 \right)^{-1} = \gamma_2^{-1} \circ \gamma_2^{-1}$ . Property (Associativity) holds for the composition of maps.
To see why (Identity) holds, consider the trivial map $\text{id}: G \to G$ . We have $\text{id} \, (a \circ b) = a \circ b = \text{id}(a) \circ \text{id}(b)$ so that $\text{id}$ is a group homomorphism, and it is clear that $\text{id}$ is a bijection. To see why (Inverses) holds, say that $g = \gamma(a)$ and $h = \gamma(b)$ . Then we have
$\begin{aligned} \gamma^{-1}( g \circ h) & = \gamma^{-1} \bigl( \gamma(a) \circ \gamma(b) \bigr) \\[5pt] & = \gamma^{-1} \bigl( \gamma(a \circ b) \bigr) \\[5pt] & = a \circ b = \gamma^{-1}(g) \circ \gamma^{-1}(h) \end{aligned}$
showing that $\gamma^{-1}$ is a also group homomorphism.

We show that $\text{ker} \, \phi$ is a subgroup of $G$ . To see why (Closure) holds, observe that $\phi(a \circ b) = \phi(a) \diamond \phi(b) = \epsilon \diamond \epsilon = \epsilon$ so that $a \circ b \in \text{ker} \, \phi$ for any $a, \, b \in \text{ker} \, \phi$ . Property (Associativity) holds because it holds for $G$ . To see why (Identity) holds, observe that $\phi(a) \diamond \phi(e) = \phi(a \circ e) = \phi(a) = \phi(a) \diamond \epsilon$ for any $a \in G$ , so that $\phi(e) = \epsilon$ . To see why (Inverses) holds, note that $\phi(a) \diamond \phi(b) = \phi(a \circ b) = \phi(e) = \epsilon$ whenever $a \circ b = e$ so that the inverse of $\phi(a)$ is $\phi(b)$ . In particular, if $\phi(a) = \epsilon$ then $\phi(a^{-1}) = \epsilon$ as well.

Finally we show that $\phi(G)$ is a subgroup of $H$ . To see why (Closure) holds, observe that $\alpha \diamond \beta = \phi(a) \diamond \phi(b) = \phi(a \circ b)$ whenever $\alpha, \, \beta \in \phi(G)$ , so that $\alpha \ast \beta \in \phi(G)$ as well. Property (Associativity) holds because it holds for $H$ . To see why (Identity) holds, recall that $\epsilon = \phi(e)$ . To see why (Inverses) holds, recall that $\phi(a^{-1}) = \phi(a)^{-1}$ . $\square$

Group Actions

Let $G$ be a group under a composition law $\circ$ with identity $e$ . A group action by $G$ on a set $V$ is a well-defined map $\circ: G \times V \to V$ , which we write as $(g, v) \mapsto g \circ v$ , satisfying the following axioms:

Associativity: $g_1 \circ \left( g_2 \circ v \right) = \left( g_1 \circ g_2 \right) \circ v$ for all $g_1, g_2 \in G$ and $v \in V$ .

Identity: $e \circ v = v$ for all $v \in V$ .

Denote the permutation group $\text{Aut}(V)$ as the set of all permutations $\gamma: V \to V$ , that is, well-defined maps which are both injective and surjective. When $G$ acts on a set $V$ , we may think of $G$ as being “contained” in $\text{Aut}(V)$ .

Proposition.
Let $G$ be a group acting on a set $V$ .

For each $g \in G$ , the map $\gamma_g: V \to V$ defined by $v \mapsto g \circ v$ is a permutation.

The map $\phi: G \to \text{Aut}(V)$ defined by $g \mapsto \gamma_g$ is a group homomorphism.

If $V = \{ v_1, \, v_2, \, \dots, \, v_n \}$ has $n$ elements, then $\text{Aut}(V)$ has $n!$ elements.

Proof: First observe that $\bigl( \gamma_g \circ \gamma_h \bigr)(v) = \gamma_g \bigl( h \circ v \bigr) = g \circ h \circ v = \gamma_{g \circ h}(v)$ for all $g, h \in G$ and $v \in V$ so that $\gamma_g \circ \gamma_h = \gamma_{g \circ h}$ . Hence $\bigl( \gamma_g \bigr)^{-1} = \gamma_{g^{-1}}$ , so that $\gamma_g$ is indeed a well-defined bijection. The map $\phi: G \to \text{Aut}(V)$ is defined by $\phi(g) = \gamma_g$ . Since $\phi(g \circ h ) = \gamma_{g \circ h} = \gamma_g \circ \gamma_h = \phi(g) \circ \phi(h)$ , we see that $\phi$ is indeed a group homomorphism.

Let $\gamma \in \text{Aut}(V)$ . Since $\gamma \circ v_1 = v_i$ , there $n$ possibilities for $i$ . By assumption, $\gamma$ is injective, so $\gamma \circ v_2 \neq \gamma \circ v_1$ . Hence there are $(n-1)$ possibilities for $\gamma \circ v_2 = v_j$ . Continuing in this fashion, we see that there are at most $n!$ possibilities for $\gamma$ . We have actually counted the number of injective maps $\gamma: V \hookrightarrow V$ . Each such map gives rearrangements for the elements of $V$ , so each is really an surjective map. Hence of these $n!$ possibilities is really a bijection. $\square$

The map $\phi: G \to \text{Aut}(V)$ defined by $g \mapsto \gamma_g$ is called the permutation representation of the associated action on $V$ . If the kernel of this group homomorphism is trivial, that is, $\text{ker} \, \phi = \{ e \}$ , then we say that $G$ acts faithfully on $V$ .

Symmetric and Alternating Groups

The symmetric group of the set $\{ 1, 2, \dots, n \}$ is simply denoted by $S_n = \text{Aut}\bigl( \{ 1, 2, \dots, n \} \bigr)$ , and is called the symmetric group of degree $n$ . This group acts on the collection $V = \{ z_1, \, z_2, \dots, \, z_n \}$ of $n$ distinct variables by $g \, z_i = z_{g(i)}$ . We will use this action to define a normal subgroup of $S_n$ .

Proposition.
The map $\text{sgn}: S_n \to \{ \pm 1 \}$ defined by $\text{sgn}(g) = \displaystyle \prod_{i < j} \bigl( z_{g(i)} - z_{g(j)} \bigr) \biggl/ \prod_{i < j} \bigl( z_i - z_j \bigr)$ is a group homomorphism.

Proof:We have the identity
$\begin{aligned} \text{sgn}(g \circ h) & = \displaystyle \prod_{i < j} \dfrac {z_{g h(i)} - z_{g h(j)}}{z_i - z_j} \\[5pt] & = \prod_{i < j} \dfrac {z_{g h(i)} - z_{g h(j)}}{z_{h(i)} - z_{h(j)}} \cdot \prod_{i < j} \dfrac {z_{h(i)} - z_{h(j)}}{z_i - z_j} \\[5pt] & = \text{sgn}(g) \cdot \text{sgn}(h) \end{aligned}$
for all $g, \, h \in S_n$ . $\square$

The image $\text{sgn}(g) = \pm 1$ is called the sign of a permutation $g \in S_n$ . We say that $g$ is even (odd, respectively) if $\text{sgn}(g) = +1$ ( $\text{sgn}(g) = -1$ , respectively). The kernel of this group homomorphism is defined as the alternating group of degree $n$ : $A_n = \text{ker} \, \text{sgn} = \left \{ g \in S_n \ \bigl| \ \text{sgn}(g) = +1 \right \}$ . Note that $A_n$ is a subgroup of $S_n$ .

Matrix Groups

We generalize the idea behind dihedral groups by considering matrices in a more general setting. Let $K$ denote either $\mathbb Q$ , $\mathbb R$ , or $\mathbb C$ . Recall that a matrix over $K$ is an $m \times n$ array, that is, a grid with $m$ rows and $n$ columns in the form $A = \left [ \begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{matrix} \right ]$ where $a_{ij} \in K$ . We will use shorthand notation to write $A = \bigl[ a_{ij} \bigr]$ to denote that $a_{ij}$ is the entry in the $i$ th row and $j$ th column. The set of all such matrices is denoted by $\text{Mat}_{m \times n}(K)$ .

We will be interested in the set square matrices over $K$ , that is, $m = n$ . The sum and product of two square matrices is defined as
$\begin{aligned} A + B & = \biggl[ \ a_{ij} + b_{ij} \ \biggr] \\[5pt] A \cdot B & = \biggl[ \ a_{i1} \, b_{1j} + a_{i2} \, b_{2j} + \cdots + a_{in} \, b_{nj} \ \biggr] \end{aligned} \qquad \text{in terms of} \qquad \begin{aligned} A & = \biggl[ \ a_{ij} \ \biggr], \\[5pt] B & = \biggl[ \ b_{ij} \ \biggr]. \end{aligned}$

Proposition.
Let $K$ denote either $\mathbb Q$ , $\mathbb R$ , or $\mathbb C$ . Then set $GL_n(K)$ consisting of those invertible $A \in \text{Mat}_{n \times n}(K)$ a group under matrix multiplication.

Proof:We have to verify four axioms. To show (Closure), note that $\left( A \cdot B \right)^{-1} = B^{-1} \cdot A^{-1}$ for any $A, B \in GL_n(K)$ so that the product of invertible square matrices is again an invertible square matrix. To show (Associativity), denote $A = \bigl[ a_{ij} \bigr]$ , $B = \bigl[ b_{ij} \bigr]$ , and $C = \bigl[ c_{ij} \bigr]$ as three elements from $GL_n(K)$ . We have
$\begin{aligned} \bigl( A \, B \bigr) \, C & = \displaystyle \biggl[ \sum_{p=1}^n a_{ip} \, b_{pj} \biggr] \, \biggl[ \ c_{ij} \ \biggr] \\[5pt] & = \biggl[ \sum_{p=1}^n \sum_{q=1}^n a_{ip} \, b_{pq} \, c_{qj} \biggr] \\[5pt] & = \biggl[ \ a_{ij} \ \biggr] \, \biggl[ \sum_{q=1}^n b_{iq} \, c_{qj} \biggr] \\[5pt] & = A \, \bigl( B \, C \bigr). \end{aligned}$
To show (Identity), note that the $n \times n$ identity matrix $I_n$ is an invertible square matrix. To show (Inverses), recall that $\left( A^{-1} \right)^{-1} = A$ , so that the inverse of an invertible square matrix is again an invertible square matrix. $\square$

The set $GL_n(K)$ is called the general linear group of degree $n$ over $K$ . When $n = 1$ , we have $GL_1(K) = K^\times = K - \{ 0 \}$ as a group under multiplication. Hence, $GL_n(K)$ is a natural generalization. There is another way to define this group. Recall that a matrix $A = \bigl[ a_{ij} \bigr]$ is invertible if and only if its determinant is nonzero: $\det \left [ \begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{matrix} \right ] = \displaystyle \sum_{g \in S_n} \text{sgn}(g) \cdot a_{1 \, g(1)} \, a_{2 \, g(2)} \, \cdots \, a_{n \, g(n)}$ . Since the determinant is a multiplicative map, that is, $\det \bigl( A \cdot B \bigr) = \bigl( \det A \bigr) \cdot \bigl( \det B \bigr)$ , we may view the determinant as a group homomorphism $\det: GL_n(K) \to GL_1(K)$ . The special linear group of degree $n$ is the kernel of this homomorphism: $SL_n(K) = \text{ker} \, \det = \left \{ A \in GL_n(K) \ \bigl | \ \text{det}(A) = 1 \right \}$ . Note that $SL_n(K)$ is a subgroup of $GL_n(K)$ .

Fields and Extensions

A triple $\bigl( F, \, +, \, \cdot \bigr)$ is said to be a field if the following three axioms hold:

Additive Group: The pair $\bigl(F, \, + \bigr)$ is an additive group with identity 0.

Multiplicative Group: The pair $\bigl(F^\times, \, \cdot \bigr)$ is a multiplicative group, where $F^\times = F - \{0\}$ .

Distributivity: For all $a, \, b, \, c \in F$ we have the distributive law $a \cdot ( b + c ) = ( a \cdot b ) + ( a \cdot c )$ .

The sets $\mathbb Q$ , $\mathbb R$ , are $\mathbb C$ are each examples of fields. A subset $K \subseteq F$ is called a subfield if $\bigl( K, \, +, \, \cdot \bigr)$ is also a field; we also say that $F$ is an extension of $K$ .

Given a variable $z$ , a polynomial over $K$ is a sum $p(z) = a_n \, z^n + \cdots + a_1 \, z + a_0$ where each $a_i \in K$ . If $a_n \neq 0$ , we say that $\deg(p) = n$ is the degree of the polynomial $p(z)$ . A rational function over $K$ is an expression $f(z) = p(z) / q(z)$ which is the ratio of two polynomials $p(z)$ and $q(z)$ over $K$ . If $p(z) = a_n \, z^n + \cdots + a_1 \, z + a_0$ and $q(z) = b_m \, z^m + \cdots + b_1 \, z + b_0$ are relatively prime, that is, the $(n+m) \times (n+m)$ determinant $\text{res}(p,q) = \det \left[ \begin{matrix} a_n & a_{n-1} & \cdots & a_0 & 0 & \cdots & 0 & 0 \\ 0 & a_n & \cdots & a_1 & a_0 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & a_n & a_{n-1} & \cdots & a_0 & 0 \\ 0 & 0 & \cdots & 0 & a_n & \cdots & a_1 & a_0 \\ b_m & b_{m-1} & \cdots & b_0 &0 & \cdots & 0 & 0 \\ 0 & b_m & \cdots & b_1 & b_0 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & b_m & b_{m-1} & \cdots & b_0 & 0 \\ 0 & 0 & \cdots & 0 & b_m & \cdots & b_1 & b_0 \\ \end{matrix} \right]$ of the Sylvester Matrix is nonzero, we say that $\deg(f) = \max \{ \deg(p), \, \deg(q) \}$ is the larger of the degrees of the polynomials in the numerator and the denominator. The collection of all such rational functions is the set $K(z) = \left \{ \left. f(z) = \dfrac {a_n \, z^n + \cdots + a_1 \, z + a_0}{b_m \, z^m + \cdots + b_1 \, z + b_0} \ \right| \ a_i, \, b_j \in K \right \}$ . It is straight-forward to verify that $\bigl( K(z), \, +, \, \cdot \bigr)$ is indeed a field and hence an extension of $K$ . We will consider those fields $F$ which are transcendental over $K$ , that is, $F$ is an extension of $K(z)$ . Note that if $J = f(z)$ is a rational function from $K(z)$ , then the collection $K(z)$ of rational functions in $z$ is an extension of the collection $K(J)$ of rational functions in $J$ . Whenever $J = f(z)$ has $\deg(f) \geq 1$ , we say that $K(z)$ is an algebraic extension of $K(J)$ of degree $\deg(f)$ .

Möbius Transformations

Let $K$ denote either $\mathbb Q$ , $\mathbb R$ , or $\mathbb C$ . The collection of rational functions $\gamma(z) \in K(z)$ which have $\deg(\gamma) = 1$ are called Möbius Transformations, named in honor of the German mathematician August Ferdinand Möbius (1790 — 1868). That is, they are in the form $\gamma(z) = \dfrac {a \, z + b}{c \, z + d}$ where $\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \in GL_2(K) = \left \{ \gamma \in \text{Mat}_{2\times 2}(K) \ \biggl| \ a \, d - b \, c \neq 0 \right \}$ . We denote this collection by $\text{Aut} \bigl( \mathbb P^1(K) \bigr)$ .

Proposition.
Let $K$ denote either $\mathbb Q$ , $\mathbb R$ , or $\mathbb C$ .

$\text{Aut} \bigl( \mathbb P^1(K) \bigr)$ forms a group under composition.

$\text{Aut} \bigl( \mathbb P^1(K) \bigr)$ acts on $K(z)$ via $\gamma^\ast \, f(z) = f \bigl( \gamma^{-1}(z) \bigr)$ .

We explain why the notation $\text{Aut} \bigl( \mathbb P^1(K) \bigr)$ . Indeed, each $\gamma$ acts on $\mathbb P^1(K)$ . Explicitly, $\tau = \bigl( \tau_1 : \tau_0 \bigr)$ maps to the Möbius transformation $\gamma(\tau) = \bigl( a \, \tau_1 + b \, \tau_0 \ : \ c \, \tau_1 + d \, \tau_0 \bigr)$ . Such a map is both injective and surjective, and hence a permutation of the projective line $\mathbb P^1(K)$ .

Proof:We explain why the $\text{Aut} \bigl( \mathbb P^1(K) \bigr)$ forms a group. Define $\circ$ as composition, that is, $\bigl( \gamma_1 \circ \gamma_2 \bigr)(z) = \dfrac {\bigl( a_1 \, a_2 + b_1 \, c_2 \bigr) \, z + \bigl( a_1 \, b_2 + b_1 \, d_2 \bigr)}{\bigl( c_1 \, a_2 + d_1 \, c_2 \bigr) \, z + \bigl( c_1 \, b_2 + d_1 \, d_2 \bigr)}$ where $\gamma_i(z) = \dfrac {a_i \, z + b_i}{c_i \, z + d_i}$ for $i = 1, \, 2$ . It is easy to check that this is a well-defined operation. In fact, this shows that the well-defined map $\phi: GL_2(K) \to \text{Aut} \bigl( \mathbb P^1(K) \bigr)$ which sends $A = \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]$ to $\gamma(z) = \dfrac {a \, z + b}{c \, z + d}$ is a surjection and preserves the composition laws: $\phi(A \cdot B) = \phi(A) \circ \phi(B)$ . But the image $\phi \bigl( GL_2(K) \bigr) = \text{Aut} \bigl( \mathbb P^1(K) \bigr)$ , so it is clear that $\text{Aut} \bigl( \mathbb P^1(K) \bigr)$ must be a group.

Consider the map $\text{Aut} \bigl( \mathbb P^1(K) \bigr) \times K(z) \to K(z)$ which sends a pair $(\gamma, \, f)$ to the rational function $\gamma^\ast \, f(z) = \bigl( f \circ \gamma^{-1} \bigr) (z) = f \left( \dfrac {d \, z - b}{-c \, z + a} \right)$ . Property (Associativity) holds because
$\begin{aligned} \gamma_1^\ast \, \bigl( \gamma_2^\ast \, f(z) \bigr) & = \gamma_1^\ast \, f \bigl( \gamma_2^{-1}(z) \bigr) \\[5pt] & = f \bigl( \gamma_2^{-1} \circ \gamma_1^{-1}(z) \bigr) \\[5pt] & = \bigl( f \circ (\gamma_1 \circ \gamma_2)^{-1} \bigr)(z) \\[5pt] & = \bigl( \gamma_1 \circ \gamma_2 \bigr)^\ast \, f(z). \end{aligned}$
for all $\gamma_1, \, \gamma_2 \in \text{Aut} \bigl( \mathbb P^1(K) \bigr)$ and $f(z) \in K(z)$ . Property (Identity) holds because the function $\text{id}(z) = z$ acts trivially: $\text{id}^\ast \, f(z) = f(z)$ . Hence $\text{Aut} \bigl( \mathbb P^1(K) \bigr) \times K(z) \to K(z)$ is indeed a group action. $\square$ .

Since Möbius transformations $\gamma: \mathbb P^1(K) \to \mathbb P^1(K)$ act on rational functions $f(z) \in K(z)$ , we define the Galois group of $f$ as the subgroup $\text{Gal}(f) = \left \{ \gamma \in \text{Aut} \bigl( \mathbb P^1(K) \bigr) \ \biggl| \ f \bigl( \gamma(z) \bigr) = f(z) \right \}$ . We are motivated by a fundamental question.

Motivating Questions.
What are the possible groups which the Galois group $\text{Gal}(f)$ can be? That is, given a finite group $G$ , can we find a rational function $f(z) \in K(z)$ such that $G \simeq \text{Gal}(f)$ ?

We will focus on these questions more in the next lecture.

2 Responses to Lecture 6: Friday, August 30, 2013

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2 Responses to Lecture 6: Friday, August 30, 2013

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