Pauli Matrices are a Group

Question

I want to show that $G= \langle X,Y,Z\rangle $ is a group of order 16, i.e, Pauli Matrices generate a group of order 16 and that the center of this group $G$ is $\langle i\cdot \operatorname{Id}\rangle$. Pauli Matrices are as follows: $$\sigma_1 \equiv \sigma_x \equiv X \equiv \begin{pmatrix} 0 && 1 \\ 1&&0 \end{pmatrix} $$ $$\sigma_2 \equiv \sigma_y \equiv Y \equiv \begin{pmatrix} 0 && -i \\ i&&0 \end{pmatrix} $$ $$\sigma_3 \equiv \sigma_z \equiv Z \equiv \begin{pmatrix} 1 && 0 \\ 0&&-1 \end{pmatrix} $$ Until now I know that $X^2=Y^2=Z^2 = \operatorname{Id}$ and I tried to show that $\langle X, Y, Z\rangle$ is a subgroup of $\operatorname{GL_n (\mathbb{C})}$. But the subgroup test doesn't seem to work since I would have to calculate for every elements $A, B$ in the group $AB^{-1}$.

I also calculated the following: $YZ = iX$, $ZY = -iX$, $ZX = iY$, $XZ = -iY$, $XY = iZ$, $YX = -iZ$, $XYZ = i\operatorname{Id}$, $XZY = -i\operatorname{Id}$, $iX\cdot i\operatorname{Id} = -X$, $iY\cdot i\operatorname{Id} = -Y$, $iZ\cdot i\operatorname{Id} = -Z$, $iX iX = - \operatorname{Id}$.

Then I have found that the 16 elements $\operatorname{Id}, X,Y,Z, -\operatorname{Id}, -X,-Y-Z, i\operatorname{Id}, iX, iY,iZ, -i\operatorname{Id},-iX,-iY,-iZ$ are in $G$. But, I do not know how to show that these are the only elements in $G$.

Answer 1

You're done.

Based on your calculations, these $16$ elements $\{i^kA:k\in\{0,1,2,3\},\, A\in\{\mathrm{Id}, X, Y, Z\}\}$ are already closed under multiplication.

Your calculation also shows that none of $X,Y,Z$ - hence neither their scalar multiples - commute every other elements, while $\lambda\mathrm{Id}$ clearly does for any $\lambda\in\Bbb C$.