Finite reflection group: Reflections with the same reflecting hyperplane are equal? Let $V$ be a finite-dimensional vector space over a field of characteristic zero.
We call a linear map $s:V \rightarrow V$ a reflection if $s^2=id_V$ and the set of its fixed points $V^s$ is a hyperplane (we call $V^s$ the reflecting hyperplane). Let $W$ be a finite reflection group (i.e. a finite subgroup of $GL(V)$ that is generated by reflections). Apparently, if two elements of $W$ have the same reflecting hyperplane they are equal. Why?
Since the reflections generate $W$ it suffices to prove the statement for the reflections. Why should they agree on $V\setminus V^s$? Somehow the finiteness of $W$ and the characteristic of the field have to enter. How?
 A: Here is a more direct argument, which avoids using Maschke's Theorem (it also works in infinite-dimensional case).
Let $s, t\in GL(V)$ be involutions fixing the same  hyperplane $H\subset V$. Then $s, t$ induce involutions of $V/H\cong F$, where $F$ is your ground field. These  induced involutions are either the identity map or the  multiplication by $-1$.  I will consider the case when both induced involutions are $-1$ since the other cases are ruled out by an argument similar to the one below (we will obtain a contradiction with the assumption that $s, t$ are involutions). Then $g=st$ induces the identity map of $F$ and fixes $H$ elementwise. Choose a vector $v\in V\setminus H$. Then there exists a vector $u\in H$ such that $g(v)=v+u$. Assume $u\ne 0$, otherwise, $g=1$ and $s=t$.
Iterating $g$ we obtain $g^k(v)=v+ ku$, $k\in {\mathbb N}$. Since $F$ has characteristic zero, $ku\ne 0$ for each $k\in {\mathbb N}$. It follows that $s, t$ generate an infinite group, which contradicts our assumptions.
