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…
 A: 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.
