About $c(A)$ in $c(A)|A|\leq |A^{-1}A|$

Question

Let $G$ be a finite group, $\emptyset\neq A\subseteq G$, $A^{-1}:=\{ a^{-1}:a\in A\}$, and put $$c(A):=\max\{t\in \mathbb{Z}: t|A|\leq |A^{-1}A|\}$$

• It is clear that $1\leq c(A)\leq \frac{|G|}{|A|}$, and if $|A|>1$ then $1\leq c(A)\leq |A|-1$.

• We observe that: if $A$ is a subgroup or $|A|=1,2$ then $c(A)=1$ (i.e., $c(A)$ takes its least possible value).

• Also, if $A^{-1}A=G$ (e.g., if $|A|>\frac{|G|}{2}$) then $c(A)$ takes its most possible value.

Now,

(1?) What are necessary and/or sufficient conditions (on $A$ and/or $G$) for being $c(A)=|A|-1$, and also for $c(A)=1$ , $c(A)=[\frac{|G|}{|A|}]$ ($[\; ]$ is the integer part notation)?

(2?) Is there any asymptotic formulas for $c(A)$ (or other lower and upper bounds)?

Answer 1

I believe that it depends on the structure of the group $G$, and the choice of it is unclear in your question.

For example in the cyclic group $G=\mathbb{Z}/n\mathbb{Z}$ you will have a have $t=1$ for the set $A=\{ 0,1,2,\ldots,k-1 \}$ if $k<2n$. Also you may take such "arithmetic progressions" in any finite group which has elements of high orders.

Also you can get such sets in another (universal) manner: Take a subgroup $H$ of $G$ and drop out $k$ elements from $H$, where $k<|H|/2$. The reduced set $A$ has at least $(|H|+1)/2$ elements, whence $AA^{-1}$ is at most size $|H|$. Hence, $t=1$.