Given an $n$, are there infinitely many finite fields $F$ such that none of the orders of $PSL_n(F)$ divide each other?

Question

Given an integer $n$, is there an infinite set of finite fields $F_i$, $i\in \mathbb{N}$ such that for $i\neq j$ we have that $|PSL_n(F_i)|$ does not divide $|PSL_n(F_j)|$.

The motivation is that then any finite group can be embedded in infinitely many finite simple groups without any of these simple groups embedding in each other.

Answer 1

This should be a comment, but as I cannot comment yet, I give it as an answer (made community wiki):

Given your motivation, it should be much easier to prove that for two finite fields $\mathbb{F}$ and $\mathbb{F'}$ of different characteristic the group $\mathop{PSL}_n(\mathbb{F})$ usually does not embed into $\mathop{PSL}_n(\mathbb{F'})$.

For this, look at the structure of $p$-Sylow subgroups for $p$ the characteristic of $\mathbb{F}$ rsp. not the characteristic of $\mathbb{F}$.

You can make your life easier by choosing the order of $\mathbb{F}^\times$ to be $\ne 1 \bmod n$ (and work with prime fields if it suits you better).

---

For instance, if G is the group you are embedding, take n to be max(|G|,3), and then for your F_i take all prime fields of size p_i congruent to 1 mod n. Then the Sylow p_i-subgroup of PSL(n,F_i) is non-abelian, while the Sylow p_j-subgroup of PSL(n,F_i) is abelian, hence PSL(n,F_i) cannot embed in PSL(n,F_j).

The important part is solely that p > n, to ensure we get Sylows contained in a maximal torus. The n ≥ 3 requirement makes the defining characteristic Sylow non-abelian, and avoids some issues when F_i is of size 4 or 5 and n is only 2, where one does get some embeddings, and when F_i is of size 2 or 3 and n is only 2 (or in general if n is 1), when we do not get simple groups.