An introduction to group representation theory

I have struggled for weeks thinking how to introduce the topic of group representation theory, and my best idea is to look at it as a game with rules. We want to define the character table of a group, which is a grid of the traces of matrices of irreducible representations.

 

For any finite group G, we can define a representation as a homomorphism h, h:G→K, where K is a group of invertible nxn matrices.

 

Stated without proof: the number of irreducible representations is equal to the number of conjugacy classes.

Stated without proof: The order of the group is equal to the sum of the squares of the dimensions of the representations.

Corollary: A finite abelian group of order n has n one-dimensional representations, and all of the entries are on the complex unit circle.

One more definition is needed: The complex conjugate of a+bi = a-bi.

Corollary #1: The complex conjugate of a real number a is a.

Corollary #2: On the complex unit circle, the multiplicative inverse of any number is its complex conjugate, (a+bi)(a-bi) = 1.

 

Let's start small, and ℤ2 is the smallest interesting case. Let's call the elements 1 and -1.

 

   | 1 | -1 

 1 |   |    

-1 |   |    

 

I will fill in the character table in red.

 

Rule #1: The homomorphism of sending everything to 1 is always an irreducible representation.

 

   | 1 | -1 

 1 | 1 |  1 

-1 |   |    

 

Rule #2: The identity is always sent to the identity, and the trace of the identity is equal to the dimension of the representation.

 

   | 1 | -1 

 1 | 1 |  1 

-1 | 1 |    

 

Rule #3: Every row of the character table can be looked upon as a vector and the dot product of one row with the complex conjugate of a different row will always be zero. The dot product of a row with itself will be the order of the group.

 

A dot product of zero indicates that the two vectors are orthogonal, which is the same as perpendicular, the two vectors meet at a 90° angle. If we are dealing in two or three dimensions, it's easy to show 90° angles, but as we get to vectors in larger dimensions, visualization becomes difficult and the dot product is our best way to determine orthogonality. 

   | 1 | -1 

 1 | 1 |  1 

-1 | 1 | -1 

Next, we look at ℤ2✕ℤ2. I will call the elements 1, a, b and ab, where the square of every element is equal to 1. 

   | 1 | a | b | ab 

 1 |   |   |    |   

 a |   |   |    |   

 b |   |   |    |   

ab |   |   |    |   

First row and first columns are all 1s.

   | 1 | a | b | ab 

 1 | 1 | 1 | 1 | 1  

 a | 1 |   |   |   

 b | 1 |   |   |   

ab | 1 |   |   |   

 

To make orthogonal vectors, we map two of the letter variables to -1, and their product to 1.

   | 1 | a | b | ab 

 1 | 1 | 1 | 1 | 1  

 a | 1 |-1 |-1 | 1  

 b | 1 |-1 | 1 |-1  

ab | 1 |-1 |-1 | 1  

One more small group to look at, ℤ4. I will call the elements 1, i, -1 and -i.

   | 1 | i | -1 | -i 

 1 |   |   |    |    

 i |   |   |    |    

-1 |   |   |    |    

-i |   |   |    |    

Yet again, all 1s in the first row and column.

   | 1 | i | -1 | -i 

 1 | 1 | 1 |  1 |  1 

 i | 1 |   |    |    

-1 | 1 |   |    |    

-i | 1 |   |    |    

Second row, send every element to itself.

 

   | 1 | i | -1 | -i 

 1 | 1 | 1 |  1 |  1 

 i | 1 | i | -1 | -i 

-1 | 1 |   |    |    

-i | 1 |   |    |    

 

Third row, send every element to its square.

 

   | 1 | i | -1 | -i 

 1 | 1 | 1 |  1 |  1 

 i | 1 | i | -1 | -i 

-1 | 1 |-1 |  1 | -1 

-i | 1 |   |    |    

 

Fourth row, send every element to its cube.

 

   | 1 | i | -1 | -i 

 1 | 1 | 1 |  1 |  1 

 i | 1 | i | -1 | -i 

-1 | 1 |-1 |  1 | -1 

-i | 1 |-i | -1 |  i 

This is the first case where we have to consider complex conjugates. The dot product of rows 2 and 4 becomes

 

1x1 + (i)(i) + (-1)(-1) + (-i)(-i) = 1 + -1 + 1 + -1 = 0.

 

In the next posts, we will look at all the groups of order 8, three of them abelian - ℤ8, ℤ4✕ℤ2 and ℤ2✕ℤ2✕ℤ2 - and two non-abelian, D4 and the quaternions.

 

A finite field with four elements

 In the last post, we discovered that for any prime p, ℤp is a finite field, where the entire set under addition is an abelian group and ℤp-{0} is an abelian group under multiplication. On the other hand, if n is a composite number, ℤn-{0} is not a group under multiplication because there will be at least two numbers between 1 and n-1 that multiply to n, which in modulo n arithmetic means j*k = 0, and neither j or k will have multiplicative inverses.

Let us instead look at an additive group of four elements, 0, 1, a and 1+a. Here is the Cayley table for addition.

 

      0    1    a    1+a

0   | 0    1    a    1+a

1   | 1    0    1+a   a 

a   | a    1+a   0    1 

1+a | 1+a  a     1    0 

Every element is its own additive inverse, so this is isomorphic to the Klein four-group. What about the non-zero elements under multiplication? There are three such elements, so if this is a group, it must be isomorphic to ℤ3. Let's figure out the Cayley table.

 

      1    a    1+a

1   | 1    a    1+a

a   | a    ?    ?  

1+a | 1+a  ?    ?  

Recall that every element must show up in every row and column, so a*(1+a) must equal 1, and likewise (1+a)*a.

      1    a    1+a

1   | 1    a    1+a

a   | a    ?    1  

1+a | 1+a  1    ?  

The other two question marks can be sussed out by using the Latin square rule.

      1    a    1+a

1   | 1    a    1+a

a   | a    1+a  1  

1+a | 1+a  1    a  

In regular multiplication, (1+a)²= 1 + 2a + a², but because this group uses mod 2 arithmetic, this simplifies to 1 + a². On the table we see a² = 1+a, so we get 1+ 1+a, and mod 2 simplifies this to a. I leave it as an exercise for the reader to see that any combination of these four elements, will be consistent with our addition and multiplication tables.

 

In the literature, a field whose order is q = pⁿ is called Fq or 𝔽q or GF(q), which stands for Galois field.

 

Finite fields

 A field F is a set that includes 0 and has two abelian operations defined, usually called multiplication and addition. F is a group under addition and F - {0} is a group under multiplication.

There are many well known infinite sets that are fields, including the real numbers ℝ, the rationals ℚ, and the complex numbers ℂ. The integers ℤ are not a field because there are no multiplicative inverses. But when we take ℤp where p is prime, this creates a finite field.

 

Let's look at ℤ5 - {0} under multiplication.

 

    1    2    3    4 

1|  1    2    3    4

2|  2    4    1    3

3|  3    1    4    2

4|  4    3    2    1

 

1 is always its own inverse, and in ℤ5 - {0}, 4 is also its own inverse. It will always be true that (n-1)²≣ 1 (mod n) for all n > 1.

 

Because 6 is not prime, ℤ6 - {0} under multiplication is not a group.

 

    1    2    3    4    5

1|  1    2    3    4    5

2|  2    4    0    2    4

3|  3    0    3    0    3    

4|  4    2    0    4    2

5|  5    4    3    2    1 

 

Because both 2 and 3 divide 6, they do not have multiplicative inverses. 4 doesn't divide 6, but they have a common factor of 2. If we want to make a multiplicative group mod 6, we have to use the numbers relatively prime to 6, which are 5 and 1.

 

    1    5

1|  1    5

5|  5    1

For any prime p, all the integers from 1 to p-1 are relatively prime, so the multiplication table will never have two non-zero numbers that multiply to 0.

There are also finite fields of the order p². In the next post, we will look at a finite field of order 4.