# Pauli Matrices are a Group

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 $$G$$.

• I have no clue of how to find the center of $G$. Nov 10, 2019 at 0:52

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 $$\lambda\in\Bbb C$$.
• The notation $\langle X,Y,Z\rangle$ denotes the smallest subgroup containing $X, Y, Z$, and these 16 elements form a subgroup, so the generated subgroup is included in there. Nov 10, 2019 at 9:34