Construction of Permutation Group (bounded order) from a Set of Permutation

Question

Given a set of permutation $S$. It has $|S|$ (The cardinality of $S$) elements. Consider a subset $A\subset S$. We can construct a group using $A$.

One possible algorithm could be-

1. We start copying elements of $A$ to $B$, so $B=A$.
2. Construct all possible product (Composition of permutations–the group product) of elements of $B$. There will be ${|B|}^{|B|}$ products.

3. If there is a new element (a new permutation), include that into set $B$ and repeat step $2$.

4. Else, there is no new element,Stop.

5. Construct $B^{-1}$ for set $B$.

6. The group is $G$,where $G=B^{-1} \cup B$ (see Generating set of a group).

Question:

Does there always exist a subset $A$ of $S$ such that $G$ is a nontrivial group and $|G| \leq |S|^{c} $ where $c$ is a constant?

Edit:

Both $|S| , |A| > $ 1 , . The summary of the question: Is there a group $G$, where $|G| \leq |S|^{c} $ , $c$ is a constant and $G$ is constructed from $S$ as described above. For example, if $c=1, then |G| \leq |S| $.