Undergrad Pin & Spin Groups: Double Covers of Orthogonal & SO Groups

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Pin Groups serve as the double cover of the Orthogonal Group, while Spin Groups serve as the double cover of the Special Orthogonal Group. Both Pin and Spin are classified as groups, but there is a distinction regarding the presence of the identity element. Only one of these double covers contains the identity element, leading to confusion about their group status. Clarification confirms that both Pin and Spin are indeed groups, despite the nuances in their definitions. Understanding these distinctions is crucial for comprehending the structure of these mathematical groups.
redtree
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Do both sets of the double cover contain an identity element?
Pin Groups are the double cover of the Orthogonal Group and Spin Groups are the double cover of the Special Orthogonal Group. Both sets of the double cover are considered to be groups, but it seems that only one of the sets of the double cover actually contains the identity element, which means that both are not groups. Am I missing something?
 
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$$
\underbrace{\{\ldots \det =\pm 1\}}_{\text{Group}}=\underbrace{\{\ldots \det =-1\}}_{\text{ no Group}}\cup \underbrace{\{\ldots \det =1\}}_{\text{Group}}
$$
 
Got it; thanks!
 
redtree said:
Got it; thanks!
Just to be clear, both Pin and Spin are groups.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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