$V$ is a finite dimensional vector space over a field $K$, $Q$ is a non-degenerate quadratic form on $V$, $Cl(V,Q)$ is the Clifford algebra.
- The Wikipedia article "Clifford algebra" says:
The pin group $Pin_V (K)$ is the subgroup of the Lipschitz group $Γ$ of elements of spinor norm 1
- The Wikipedia article "Clifford algebra" and "Pin group" says:
The pin group $Pin_V (K)$ is the subset of the Clifford algebra $Cl(V,Q)$ consisting of elements of the form $v_1 v_2 \cdots v_k$, where the $v_i$ are vectors such that $Q(v_i)=\pm 1$
I think $1$ is the best definition since it is directly related to the exact sequence $1\to{\pm 1}\to Pin_V (K)\to O_V(K)\to K^{\times}/{K^{\times}}^2$.
And $2$ is better to be regarded as a property of the pin group.
But how to prove $2$ is equivalent to $1$? At least in the process to prove $2\to 1$ I have such problems:
First according to $1$, spinor norm here is defined as $x^t x$ where $x$ is any element of $Cl(V,Q)$, and for elements from $Γ$ (as a subset of $Cl(V,Q)$) the spinor norm will be in $K^{\times}$ (as the grade $0$ vector subspace of $Cl(V,Q)$).
Also according to $1$, for elements $x$ from $V$ (as the grade $1$ vector subspace of $Cl(V,Q)$), $x^t =x$.
Also since $v_i$ is from $V$, then ${v_i}^2=Q(v_i)$.
So the spinor norm of elements of the form described by $2$ is:
$$(v_1 v_2 \cdots v_k)^t(v_1 v_2 \cdots v_k)=({v_k}^t \cdots {v_1}^t)(v_1 v_2 \cdots v_k)=(v_k \cdots v_1)(v_1 v_2 \cdots v_k)=Q(v_1) \cdots Q(v_k)=\pm 1$$
So the spinor norm is $\pm 1$, not the same as $1$ describes.