We consider $L=\ \{x\in (\mathbb{Z}/2^3 \mathbb{Z})^{\times} \ \vert \ x^2 \equiv 1 \pmod{2^3}\}$. I want to determine $Card(L)$.
If the group was the cyclic the answer would be $Card(L)=\gcd(4,2)=2$ but it's not the case here because $(\mathbb{Z}/2^3 \mathbb{Z})^{\times}$ isn't cyclic.
Or if we consider that $2^3=8$ we can try to solve the congruence $x^2\equiv 1 \pmod 8$ testing every number from $(\mathbb{Z}/8\mathbb{Z})^{\times}$. But we have the isomorphism with $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ so the four elements are : $(0,0),(1,0),(0,1),(1,1)$ but how can we proceed to conclude ?
Thanks in advance !