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.

Facts used

 * 1) uses::Splitting criterion for conjugacy classes in the special linear group
 * 2) uses::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: