Embedding of finite groups It is well known that any finite group can be embedded in Symmetric group $S_n$, $GL(n,q)$ ($q=p^m$) for some $m,n,q\in \mathbb{N}$. Can we embed any finite group in $A_n$, or $SL(n,q)$ for some $n,q\in \mathbb{N}$?

Yes.
The symmetric group $Sym(n)$ is generated by $\{(1,2), (2,3),\ldots, (n−1,n)\}$.  You can embed $Sym(n)$ into $Alt(n+2)$ as the group generated by $\{(1,2)(n+1,n+2), (2,3)(n+1,n+2), …, (n−1,n)(n+1,n+2)\}$.  This embedding takes a permution $\pi\in Sym(n)$ and sends it to $\pi⋅(n+1,n+2)^{\text{sgn}(\pi)}$, where $\text{sgn}(\pi)\in\{0,1\}$ is the parity of the permutation.
In other words, $G\le Sym(n)\le Alt(n+2)$ embeds any group into a (slightly larger) alternating group.
The general linear group $GL(n,q)$ embeds in the special linear group $SL(n+1,q)$ using a determinant trick.  We just add a new coordinate to cancel out the determinant of the matrix from $GL(n,q)$ so the result lands in $SL(n+1,q)$.
$$\operatorname{GL}(n,q) \cong \left\{ \begin{bmatrix} A & 0 \\ 0 & 1/\det(A) \end{bmatrix} : A \in \operatorname{GL}(n,q) \right\} ≤ \operatorname{SL}(n+1,q)$$
In other words, $G\le GL(n,q)\le SL(n+1,q)$ embeds any group into a (slightly larger) special linear group.
