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

## 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.

## 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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