Minimal possible order of a group that contains a specific subset Suppose $G$ is a group. Suppose $A \subset G$ is a subset of $G$ satisfying the following condition: $\forall a \in A \exists ! b \in A$ such that $[a, b] \neq e$. Suppose $|A| = 2n$. What is the minimal possible order of $G$?
I can build such group of order $2^{2n+1}$, namely $G = \langle a_1, … , a_n, b_1, … , b_n, c| a_i^2 = b_i^2 = c^2 = [a_i, c]=[b_i, c]=[b_i,b_j] = [a_i, a_j] = e, [a_i, b_j] = c^{\delta_{ij}}\rangle$, where $\delta$ stands for the Kronecker delta function, and $A = \{a_1, … ,a_n, b_1, … , b_n\}$.
However, I do not know, whether $2^{2n+1}$ is the minimal possible order, or is there some  better construction…

I've been thinking about the same question, in relation to your previous question. (I think you should link to it to explain your motivation.)
The short answer is no, as for $n=1$ we can take $G=S_3$ and $A$ a pair of involutions. But I think it may be true for $n\geq 2$.
EDIT: Here's a proof, by induction on $n$.
We start with the base case, $n=2$. To ease the notation, I'll write $A=\{a,b,x,y\}$, where $[a,b]\neq 1\neq [x,y]$ (with the others commuting).  I'll also write $C_a$ for the centraliser of $a$ in $G$, and so on.  
Clearly, we can assume that $G=\langle A\rangle$. Note that $C_a\cap C_b$ is a nonabelian group (since it contains the noncommuting elements $x$ and $y$) so $|C_a\cap C_b|\geq 6$. Similarly  $|C_x\cap C_y|\geq 6$. If $Z(G)=1$, then $(C_a\cap C_b)\cap (C_x\cap C_y)=1$ and so $|G|\geq |C_a\cap C_b||C_x\cap C_y|\geq 36$. 
We can therefore assume that $Z(G)\neq 1$. This implies that $C_a\cap C_b$ is a nonabelian group with nontrivial center, so $|C_a\cap C_b|\geq 8$. Now, $a\in C_a\setminus (C_a\cap C_b)$ and  $b\in G\setminus C_a$, so $C_a\cap C_b<C_a<G$. It follows that $|G|\geq 4|C_a\cap C_b|\geq 32$.
Finally, the induction step: assume $n\geq 3$ and that the result is true for $n-1$. Remove a pair of generators $a$ and $b$, to obtain $A'$ and $G':=\langle A'\rangle$. By induction $|G'|\geq 2^{2n-1}$. Now, $G'\leq C_a\cap C_b<C_a<G$, for the same reasons as above, so $|G|\geq 4|G'|\geq 2^{2n+1}$, closing the induction.
