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.

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.

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).

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.

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 ith row and jth 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}

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).

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.

About Edray Herber Goins, Ph.D.

Edray Herber Goins grew up in South Los Angeles, California. A product of the Los Angeles Unified (LAUSD) public school system, Dr. Goins attended the California Institute of Technology, where he majored in mathematics and physics, and earned his doctorate in mathematics from Stanford University. Dr. Goins is currently an Associate Professor of Mathematics at Purdue University in West Lafayette, Indiana. He works in the field of number theory, as it pertains to the intersection of representation theory and algebraic geometry.
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2 Responses to Lecture 6: Friday, August 30, 2013

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