My goal is to identify the $Pin$ group $$ 1 \to Spin(n) \to Pin^{\pm}(n) \to \mathbb{Z}_2 \to 1 $$ such that $Pin^{\pm}(n)$ are isomorphisms to other more familiar groups.
My trick is that to look at the center $Z$ of $Pin$ group, say $Z(Pin(n))$.
We see that, for $k \in \mathbb{Z}^+$,
when $n=4k+1$, $$ Z(Pin^+(4k+1))=\mathbb{Z}_2 \times \mathbb{Z}_2, $$ $$ Z(Pin^-(4k+1))=\mathbb{Z}_4. $$
when $n=4k+3$, $$ Z(Pin^+(4k+3))=\mathbb{Z}_4, $$ $$ Z(Pin^-(4k+3))=\mathbb{Z}_2 \times \mathbb{Z}_2. $$
when $n=4k$ or when $n=4k+2$, $$ Z(Pin^+(n))=Z(Pin^-(n))=\mathbb{Z}_2. $$
So my naive attempt is that we have the following $Pin$ group isomorphisms:
$$Pin^+(1)=\mathbb{Z}_2 \times \mathbb{Z}_2, \quad Pin^-(1)=\mathbb{Z}_4.$$
$$Pin^+(3)=SO(3) \times \mathbb{Z}_ 4, \quad Pin^-(3)=Spin(3) \times \mathbb{Z}_2.$$
$$Pin^+(5)=Spin(5) \times \mathbb{Z}_2,\quad Pin^-(5)=SO(5) \times \mathbb{Z}_4.$$
$$Pin^+(4k+1)=Spin(4k+1) \times \mathbb{Z}_2,\quad Pin^-(4k+1)=SO(4k+1) \times \mathbb{Z}_4.$$
$$Pin^+(4k+3)=SO(4k+3) \times \mathbb{Z}_ 4, \quad Pin^-(4k+3)=Spin(4k+3) \times \mathbb{Z}_2.$$
Question 1: Am I correct about the above statements? I am not so sure, $Pin^+(3)=SO(3) \times \mathbb{Z}_ 4$, could it be that instead $Pin^+(3)=\frac{Spin(3) \times \mathbb{Z}_4}{\mathbb{Z}_2}?$ (It looks that mine agrees with Wikipedia, but I doubt Wikipedia may be wrong about $Pin^+(3)=SO(3) \times \mathbb{Z}_ 4$...)
Question 2: Are there other $Pin$ group isomorphisms for $n=4k$, or $4k+2$, say $$Pin^{+/-}(4k)\simeq ?$$ and $$Pin^{+/-}(4k+2)\simeq?$$
Thanks for your comments in advance.