Finding nice relations for an explicit matrix group and showing that it is isomorphic to the symmetric group

Question

Let $A$ be the $n \times n$ matrix with all diagonal entries equal to 1 except in the last column and all entries in the last column equal to -1 and all other entries equal to 0. Let $B$ be the $n \times n$ matrix corresponding to the permutation sending $i$ to $i+1$ (and $n$ to $1$). For example for $n=4$, the matrices look (in sage) as follows:

A=matrix([[1,0,0,-1],[0,1,0,-1],[0,0,1,-1],[0,0,0,-1]]) B=matrix([[0,0,0,1],[1,0,0,0],[0,1,0,0],[0,0,1,0]]) 

Question: What is the group $G$ generated by the matrices $A$ and $B$ in terms of generators and (hopefully nice) relations and what is the isomorphism type of $G$?

I conjecture that it is the symmetric group $S_{n+1}$. This is true for $n \leq 10$ using Sage. But I do not see nice relations. We always have the relations $A^2=1, B^{n}=1, (BA)^{n+1}=1$, but I do not see other nice relations so far. Here is the description via Sage for $n=4$:

A=matrix([[1,0,0,-1],[0,1,0,-1],[0,0,1,-1],[0,0,0,-1]]) B=matrix([[0,0,0,1],[1,0,0,0],[0,1,0,0],[0,0,1,0]]) G = MatrixGroup([A,B]) display(len(G)) display(G.as_permutation_group().structure_description()) display(G.as_permutation_group().as_finitely_presented_group(reduced=True)) 

Result:

Finitely presented group < a, b | a^2, b^4, (a*b^-1)^5, b*a*b*(b*a*b^-2*a)^2*b^-2*a > 

Here an attempt for a special case:

When $p+1$ is a prime, then maybe one can somehow show that the group embedds into $S_{p+1}$ and then we know it must be the full symmetric group as it has an element of order 2 and p+1.

Answer 1

Consider the $n$-dimensional reduced permutation representation of $\Sigma_{n+1}$, spanned by $v_1,\ldots,v_{n+1}$ subject to $\sum v_i=0$, with $\Sigma_{n+1}$ acting by permuting the $v_i$. We take $v_1,\ldots,v_n$ as basis. Then:

• the transposition $(n\;\;(n+1))$ acts by the matrix $A$.
• the cycle $(1\;\;2\ldots n)$ acts by $B$.

These two cycles are also easily seen to generate $\Sigma_{n+1}$. So your group agrees with the image of $\Sigma_{n+1}\to \mathrm{GL}_n$. But this is a faithful representation, so this image is $\Sigma_{n+1}$.