# Why is clifford group a group?

Let $$C(Q)$$ denote the clifford algebra of vector space $$Q$$ with respect to a quadratic form $$q:V \rightarrow \Bbb R$$. Hence we have the relation $$w^2 = Q(w) \cdot 1$$ for $$w \in V$$.

Let $$\alpha:C(Q) \rightarrow C(Q)$$ be the canonical automoprhism $$\alpha^2=id, \alpha=-x$$.

The Clifford group of $$Q$$ is $$\Gamma (Q) = \{ x \in C(Q)^* \, ; |, \alpha(x) \cdot v \cdot x^{-1} \in V \text{ for all } v \in V \}$$

How is this set closed under inverses?

• You can rewrite the condition as $\alpha(x)Vx^{-1}=V$. – Lord Shark the Unknown Feb 12 at 5:39
• Ok it doesn't seem clear to me how the equality holds: the way is define $f_x :V \rightarrow V$, then as $\alpha(x) \cdot v \cdot x^{-1}$. This map is injective, $V$ finite dimensinoal so bijective. Is there an easier way? – CL. Feb 12 at 5:51