We cannot embed $F_q^\times $ in $S_n$ for small $n$ if $q − 1$ has “large” (I know it is vague) prime factors.
$F_q^\times $ is a field where $q = p^t$, where $p$ is a prime.
It is easy to see that we can always embed $F_q^\times $ in $S_t$ for "large " value of $t$
Question : What is a relation between $q$ and $t$ so that we can always embed $F_q^\times $ in $S_t$?
I tried to search on google but did not find out anything.
It's a standard theorem that $F_q^\times$ is cyclic of order $q-1$. Embeddings from cyclic groups are easy to describe: For a cyclic group $C$ of order $n$ and any group $G$, we can embed $C$ into $G$ if and only if $G$ has an element of order $n$.
To determine the possible orders in the symmetric group $S_n$, note that any element in $S_n$ may be decomposed into disjoint cycles whose cycle lengths sum to $n$ and the order of that element is the least common multiple of the cycle lengths.
So with this in mind, we can say that $F_q^\times$ embeds into $S_n$ if and only if there are natural numbers $n_1, \dots n_k$ such that $\sum_{i=1}^k n_i=n$ and $\operatorname{lcm}(n_1, \dots, n_k)=q-1$.
From this, you can see in particular that it is impossible to embed $F_q^\times$ into $S_n$ if $q-1$ has a prime factor larger than $n$.