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

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.

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Do you have a formula for the order of these groups? – Gerry Myerson Jan 25 '12 at 22:35
Given that motivation, don't you only need this to be true for some specific $n \ge 2$? – Qiaochu Yuan Jan 25 '12 at 22:36
Well, I just need it to be true for large enough $n$, as otherwise, we might not have the group embedded in the $PSL_n$. Just from the group perspective, one could indeed require quite a lot less than I ask. – Tobias Kildetoft Jan 25 '12 at 22:57

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 Fi take all prime fields of size pi congruent to 1 mod n. Then the Sylow pi-subgroup of PSL(n,Fi) is non-abelian, while the Sylow pj-subgroup of PSL(n,Fi) is abelian, hence PSL(n,Fi) cannot embed in PSL(n,Fj).

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 Fi is of size 4 or 5 and n is only 2, where one does get some embeddings, and when Fi 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.

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I specialized your very good comment to an answer. As long as the field is big compared to the dimension, and the dimension is not too small, we can just check "abelian Sylow" versus "nonabelian Sylow" and be fine. Your comments on making life easier are definitely true if one was taking smaller fields or dimension 2. – Jack Schmidt Jan 26 '12 at 18:50
This certainly makes things easier. One thing I am uncertain of (a very small detail). Can we always embed a group of order $n$ in $PSL_n(F)$? I could only find an argument for how to embed it into $PSL_{n+1}(F)$ – Tobias Kildetoft Jan 26 '12 at 20:58
@Tobias: I think so, but I'm not completely sure. A group of order $n$ acts (regularly) on $n$ points. If you take these points to be the basis of an $n$-dimensional vector space over the field $\mathbb{F}$, the linear action of $G$ fixes the $1$-dimensional subspace generated by the sum of all $n$ basis vectors. If you take the quotient of the $n$-dimensional vector space by this one-dimensional subspace, you get an embedding of your group of order $n(>1)$ into $\mathop{GL}_{n-1}(\mathbb{F})$. Probably you can build your embedding using this, but I think it's worth another question. – ego Jan 30 '12 at 17:56