Let be the field of integers modulo , with a prime. We will construct some of the irreducible representation of by induction from the subgroup of upper triangular matrices. We will use Mackey theory to show that these representations are irreducible.
The order of the group is
We let be the subgroup of upper triangular invertible matrices:
The group is sometimes called a Borel subgroup.
The index of in is . A set of left coset representatives for is
We write . We will want to know the double cosets of in , and the result is
We have
For later use, we compute
so that
Let and be two characters of . Then there is a character
defined by
Define
If is a character, then we get a character
taking to .
If are characters, then
On , we have
If are characters, then:
If , then
is an irreducible representation of dimension .
There is an irreducible representation of dimension such that, for any ,
There is an isomorphism
if and only if or .
For , the representations and are not isomorphic.
One can prove the theorem using Mackey theory, but it is also possible to avoid that and directly compute the inner products of the induced characters.
Finally, we can compute the character of .
We will need to know the conjugacy classes of and of . The characteristic polynomial almost determines the conjugacy classes; the result is as follows. Recall that the size of the conjugacy class of is where is the centralizer of in .
Every element of is conjugate to exactly one of the following elements:
such that has no solutions in .
For each type of element , the order of the centralizer subgroup is:
.
We sketch the proof. Let be the characteristic polynomial of an element . If it has distinct roots in , then is diagonalizable and we are in case 3. If it has no roots in , then choosing a basis (for some ) puts us in case 4. If the roots are identical, both equal to , there are two possibilities: either the matrix is a scalar matrix, giving case 1, or is a basis for some and then we are in case 2.
In each case, the centralizer can be computed explicitly and checked to have the order claimed (the most difficult case is the fourth). ∎
Now let be a character. We compute the character of ; after twisting by a character of the form this gives the general case. In the case , we also record the character of .
The characters are as follows (where , and has no solutions in ):