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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…

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  • $\begingroup$ It's confusing to use $e$ to denote the identity if you are using $a,b,c$ as group elements. It is standard to use $1$ for the identity in group presentations. $\endgroup$
    – Derek Holt
    Oct 15 '19 at 21:39
  • $\begingroup$ For context, you may also note that this is an extraspecial group, so quite well-known. (Well, you're actually missing some relations, namely that $c$ commutes with the $b_i$'s and the $b_i$s commute with each other.) $\endgroup$
    – verret
    Oct 15 '19 at 23:01
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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.

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