# 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…

• 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. Oct 15 '19 at 21:39
• 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.) Oct 15 '19 at 23:01

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 $$C_a\cap C_b. It follows that $$|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, $$G'\leq C_a\cap C_b, for the same reasons as above, so $$|G|\geq 4|G'|\geq 2^{2n+1}$$, closing the induction.