Presentation of a group generated by reflections through hyperplane

Question: Let $$P_i\in\mathbb{R}^n$$ be the hyperplane $$x_i - x_{i+1} = 0$$. Find a presentation for the group $$G$$ generated by the reflections in $$P_1, \ldots, P_{n-1}$$

Attempt: I really don't know how to start with this problem, which was introduced at the same time as group presentation notation. I understand that the determinant of a reflection through a hyperplane is $$-1$$, but then doesn't this mean that the composition of two such reflections would have determinant $$1$$, and hence not be a reflection? I would appreciate any hints as to how I could start this problem.

• Yes, the product of two reflections is not a reflection, but that is not an issue as you are finding the group generated by the reflections. Possibly you should start by working out: what are the orders of the generators? – user1729 Mar 20 at 11:50
• In co-ordinate terms, $P_i$ switches the $i$ and $i+1$ co-ordinates, so $(\dots, x_i, x_{i+1}, \dots) \mapsto (\dots, x_{i+1}, x_i, \dots)$. This point of view might help you see which pairs of generators commute. – gandalf61 Mar 20 at 14:38
• I guess there's some context (Coxeter groups?) in which such an exercise has been given... – YCor Mar 20 at 14:42
• @Andrew If $n \ge 4$ then there are other pairs of commuting generators. For example, $P_1$ switches first and second co-ordinates and $P_3$ switches third and fourth co-ordinates so $P_1$ will commute with $P_3$. – gandalf61 Mar 21 at 9:57
• Yes, the generators have order two. You can therefore add $P_i^2=1$ as a relation, as well as the commutators which gandalf61 pointed out. Your task is then to determine if these are all the relations or do you need more? If this is indeed all of them, then you need to prove this. – user1729 Mar 21 at 10:15

Each $$P_i$$ is a linear transformation, and a linear transformation of $$\mathbb R^n$$ is determined by what it does to the standard basis $$e_1,\ldots,e_n$$. Furthermore, each $$P_i$$ permutes the set $$\{e_1,...,e_n\}$$, because it interchanges $$e_i$$ with $$e_{i+1}$$ and fixes all the other $$e$$'s. And then, since the list $$e_1,...,e_n$$ has no repetitions, you can even simplify the notation further and think of $$P_i$$ as a permutation of the index set $$\{1,...,n\}$$. This makes the group generated by the $$P_i$$'s isomorphic to a subgroup of the permutation group $$S_n$$ on the symbols $$\{1,...,n\}$$, where $$P_i$$ itself corresponds to the permutation that, written in cycle form, is $$P_i = (i,i+1)$$.
So if you know anything about the permutation group $$S_n$$, then you might be in a very good position to continue on with this problem......