A symmetric-like group and the quaternion group $Q_8$

Question

It is well known that the symmetric group $S_n$ admits presentation with $\{(ij) \mid i\neq j\}$ as the set of generators and the following list of relations (in every formula distinct letters denote distinct indices): \begin{align} (ij) =&\, (ji), \label{Sym0} \tag{S0} \\ (ij)^2 = &\, 1, \label{Sym1} \tag{S1} \\ (jk)(ij)(jk) = &\, (ik), \label{Sym2} \tag{S2} \\ [(ij), (kl)] = &\,1. \label{Sym3} \tag{S3} \end{align}

If for $n\geq 3$ we drop relation \ref{Sym0} from this list, we will get an extension of $S_n$, denote it by $\widetilde{S}_n$. My question is: is there a standard name for this group, has it already been studied by anyone in any context? Is there any standard representation for it.

Using GAP I was able to compute $\widetilde{S_3}$ and $\widetilde{S_4}$ explicitly, the answers are [48, 41]; [384, 20069], respectively. So it looks like $\widetilde{S}_n$ is a nontrivial extension of $S_n$ by some finite nonabelian group of order $2^n$. As for the kernels $K_n=\mathrm{Ker}(\widetilde{S}_n\to S_n)$ the answers for $n=3,4,5$ are as follows: [8, 4], [16, 12], [32, 50]. In other words, $K_3 \cong Q_8$, $K_4 \cong Q_8 \times C_2$, and $K_5$ is isomorphic to the central product of $Q_8$ and $D_8$ over $C_2$.

Answer 1

To simplify typing I will call these groups $G_n$ rather than $\tilde{S}_n$. Note that $G_2$ is just the free product of two groups of order $2$, so is the infinite dihedral group, and I conjecture that $|G_n| = 2^n n!$ for $n \ge 3$, and $G_n$ is a central product of an extraspecial or symplectic type $2$-group of order $2^n$ with a double cover $2.S_n$ of $S_n$.

I believe that computations within the group $G_3$ provide enough information to prove that $|G| \le 2^nn!$ and, since $G$ maps onto $S_n$ and onto an elementary abelian group of order $2^n$, we have $|G_n| \ge 2^{n-1}n!$, so I have not quite completed the proof. I have checked the conjecture computionally for $n \ge 7$. The alternative would be that $G_n$ is the direct product of elementary abelian $2^{n-1}$ and $S_n$. I cannot see at the moment how to rule that out.

Let $H_{n-1}$ be the image of $G_{n-1}$ in $G_n$ under the map that maps each generator of $H_{n-1}$ to the generator of $G_n$ with the same name. We do not know a priori that this map is injective, and indeed it is not injective for $n=3$. We prove that $|G_n| \le 2^nn!$ by showing that $|G_n:H_{n-1}| \le 2n$ for all $n$.

We claim that the set $$T = \{ 1,\,(1n)(n1)\} \cup \{(in),\, (ni) : 1 \le i \le n-1\}$$ contains a set of right coset representatives of $H_{n-1}$ in $G_n$ and, since $|T|=2n$, this will prove that $|G_n:H_{n-1}| \le 2n$.

We can check that directly computationally (in GAP or Magma) for $n=3$, and also that $(31)(13)$, $(23)(32)$ and $(32)(23)$ all lie in the same coset as $(13)(31)$.

To prove the claim for general $n$, it is enough to show that, for each $t \in T$ and each generator $x$ of $G_n$, $tx$ lies in $H_{n-1}u$ for some $u \in T$. In fact this follows from the calculations in the $n=3$ case together with the commuting relations of $G$. This is because $t$ and $x$ either involve at most $3$ of $1,2,\ldots,n$, in which case it follows from the $n=3$ case, or they involve $4$, in which case $x \in H_{n-1}$ and $tx=xt$.

For example, when $n=5$ if $t=(25)$ and $x=(35)$ then $tx \in H_{n-1}(25)$ from the $n=3$ case applied with $1,2,3$ replaced by $2,3,5$.