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…

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.