Does reflection group induce group structure on $\mathbb{R}P^n$

Question

Let $\text{Ref}_a: \mathbb{R}^n \to \mathbb{R}^n$ be a reflection through the hyperplane defined by $a \in \mathbb{R}^n$. It seems obvious to me that $\text{Ref}_a\circ \text{Ref}_b=\text{Ref}_c$ for some $c \in \mathbb{R}^n$ (since for every two points $x,y \in S^n$ we find a hyperplane $H$, s.t. $d(H,x)=d(H,y)$). This seems a bit naive, but is there an associative multiplication $\#$ on $\mathbb{R}P^n$, s.t. $\text{Ref}_a\circ \text{Ref}_b=\text{Ref}_{a\#b}$?

Answer 1

A reflection through a hyperplane has determinant $-1$. Hence the composition of two such reflections has determinant $1$, therefore it is not a reflection through a hyperplane. So what may seem obvious is in fact false.

With respect to composition, the reflections through hyperplanes generate the unitary group. It consists of all reflections and rotations of $\Bbb{R}^n$. Its action on $\Bbb{R}^n$ descends to an action on $\Bbb{R}P^n$, leading to the projective unitary group. It is in general much larger than $\Bbb{R}P^n$ itself, which makes it unlikely to give rise to any kind of multiplication on $\Bbb{R}P^n$ (canonically).