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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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    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.2012-01-25

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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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    @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.2012-01-30