In general, for the groups in a family of fixed Lie rank, there will be a bound on the degree of an alternating group that can occur as a subgroup, so the answer to your question is no. This is easily seen from the fact that they have representations of a fixed degree. For example $E_8(q)$ has a representation of degree $248$ over ${\mathbb F}_q$, so it cannot possibly contain $A_n$ for $n > 250$, and I would guess that there is a much lower bound than that.

Of course, if by a family you mean one of the doubly infinite families like $A_n(q)$ for arbitary $n$ then the acswer is yes, because, for each such family, by making $n$ sufficient large, the groups will contain alternating groups of arbitarily large degrees as subgroup of their Weyl groups.

To be more specific, the image of the natural permutation representation of $A_n$ over ${\mathbb F}_q$ preserves a unitary form and an orthogonal form with matrix $I_n$, so $A_n \le L_n(q)$, $A_n \le U_n(q)$ and $A_n$ lies in one of the types of orthogonal groups. I am not sure if it lies in the orthogonal type that preserves a diagonal form with non-square determinant, but that type certainly contains $A_{n-1}$. It is also easy to see that $A_n \le {\rm Sp}_{2n}(q)$ for all $q$. That deals with all of the doubly infinite families.

In general, for the groups in a family of fixed Lie rank, there will be a bound on the degree of an alternating group that can occur as a subgroup, so the answer to your question is no. This is easily seen from the fact that they have representations of a fixed degree. For example $E_8(q)$ has a representation of degree $248$ over ${\mathbb F}_q$, so it cannot possibly contain $A_n$ for $n > 250$, and I would guess that there is a much lower bound than that.

Of course, if by a family you mean one of the doubly infinite families like $A_n(q)$ for arbitrary $n$ then the answer is yes, because, for each such family, by making $n$ sufficient large, the groups will contain alternating groups of arbitrarily large degrees as subgroup of their Weyl groups.

To be more specific, the image of the natural permutation representation of $A_n$ over ${\mathbb F}_q$ preserves a unitary form and an orthogonal form with matrix $I_n$, so $A_n < L_n(q)$, $A_n < U_n(q)$ and $A_n$ lies in one of the types of orthogonal groups. I am not sure if it lies in the orthogonal type that preserves a diagonal form with non-square determinant (but I think it does), but that type certainly contains $A_{n-1}$. It is also easy to see that $A_n < {\rm Sp}_{2n}(q)$ for all $q$. That deals with all of the doubly infinite families.

In general, for the groups in a family of fixed Lie rank, there will be a bound on the degree of an alternating group that can occur as a subgroup, so the answer to your question is no. This is easily seen from the fact that they have representations of a fixed degree. For example $E_8(q)$ has a representation of degree $248$ over ${\mathbb F}_q$, so it cannot possibly contain $A_n$ for $n > 250$, and I would guess that there is a much lower bound than that.

Of course, if by a family you mean one of the doubly infinite families like $A_n(q)$ for arbitary $n$ then the acswer is yes, because, for each such family, by making $n$ sufficient large, the groups will contain alternating groups of arbitarily large degrees as subgroup of their Weyl groups.

In general, for the groups in a family of fixed Lie rank, there will be a bound on the degree of an alternating group that can occur as a subgroup, so the answer to your question is no. This is easily seen from the fact that they have representations of a fixed degree. For example $E_8(q)$ has a representation of degree $248$ over ${\mathbb F}_q$, so it cannot possibly contain $A_n$ for $n > 250$, and I would guess that there is a much lower bound than that.

Of course, if by a family you mean one of the doubly infinite families like $A_n(q)$ for arbitary $n$ then the acswer is yes, because, for each such family, by making $n$ sufficient large, the groups will contain alternating groups of arbitarily large degrees as subgroup of their Weyl groups.

To be more specific, the image of the natural permutation representation of $A_n$ over ${\mathbb F}_q$ preserves a unitary form and an orthogonal form with matrix $I_n$, so $A_n \le L_n(q)$, $A_n \le U_n(q)$ and $A_n$ lies in one of the types of orthogonal groups. I am not sure if it lies in the orthogonal type that preserves a diagonal form with non-square determinant, but that type certainly contains $A_{n-1}$. It is also easy to see that $A_n \le {\rm Sp}_{2n}(q)$ for all $q$. That deals with all of the doubly infinite families.