Conjugacy class of elements with semisimple generalized Jordan block does not split in special linear group over a finite field

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Statement

Suppose q is a prime power and n is a natural number. Then, there is a finite field \mathbb{F}_q of size q, unique up to isomorphism. We denote the special linear group SL(n,\mathbb{F}_q) by SL(n,q). Similarly, we denote the general linear group GL(n,\mathbb{F}_q) by GL(n,q).

Suppose g is an element of the special linear group SL(n,q) and hence also of the general linear group GL(n,q). Suppose that, when g is written in generalized Jordan block form, there is at least one generalized Jordan block that is semisimple, i.e., there are no repeated eigenvalues within the block. Then, the GL(n,q)-conjugacy class of g is precisely the same as the SL(n,q)-conjugacy class of g. In other words, the GL(n,q)-conjugacy class of g does not split in SL(n,q).

Note that the result applies in particular to the case where g itself is semisimple (i.e., all its generalized Jordan blocks are semisimple) but it also applies in many cases where g is not semisimple.

Facts used

  1. Splitting criterion for conjugacy classes in the special linear group
  2. Norm map is surjective for finite fields

Proof

Given: Natural number n, prime power q, element g of SL(n,q) whose generalized Jordan canonical form has a semisimple generalized Jordan block

To prove: The conjugacy class of g in GL(n,q) does not split in SL(n,q).

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Without loss of generality, we can assume g itself to be a generalized Jordan form matrix (think about it)
2 The centralizer of g in GL(n,q) contains the internal direct product of the centralizers of each of the Jordan blocks. (think about it)
3 The image of the centralizer of g in GL(n,q) under the determinant map contains the subgroup generated by the images (under the determinant map) of the centralizers of each of the Jordan blocks in their respective-sized matrix groups Step (2) (think about it)
4 If there is a semisimple generalized Jordan block, the image of its centralizer under the determinant map is all of \mathbb{F}_q^\ast. Fact (2) Follows because the norm map is the same as the determinant map (elaborate, link to additional fact eventually)
5 The image of the centralizer of g in GL(n,q) under the determinant map is all of \mathbb{F}_q^\ast g has a semisimple generalized Jordan block Steps (3), (4) Step-combination direct
6 The GL(n,q)-conjugacy class of g does not split in SL(n,q) Fact (1) Step (5) Step-fact combination direct
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