Pauli product and adjoint

On one factor, $Z^{z_j}{X}^{x_j^{\mathrm{\prime }}}=(-1{)}^{z_j{x}_j^{\mathrm{\prime }}}{X}^{x_j^{\mathrm{\prime }}}{Z}^{z_j}$, since the factors commute unless both exponents are $1$, when $ZX=-XZ$ by [F1]. Multiplying these identities over the ordered factors gives $Z^z{X}^{x^{\mathrm{\prime }}}=(-1{)}^p{X}^{x^{\mathrm{\prime }}}{Z}^z$, where $p$ is the integer representative $0$ or $1$ of $z\cdot x^{\mathrm{\prime }}\in {\mathbb{F}}_2$.

Substituting that reordering into $i^{k+k^{\mathrm{\prime }}}{X}^x{Z}^z{X}^{x^{\mathrm{\prime }}}{Z}^{z^{\mathrm{\prime }}}$ gives $i^{k+k^{\mathrm{\prime }}+2p}{X}^{x+x^{\mathrm{\prime }}}{Z}^{z+z^{\mathrm{\prime }}}$, which is exactly the stated product formula; all binary exponents add modulo $2$, while the phase exponent adds modulo $4$.

The formula proves closure. $P(0,0,0)=I$ is an identity, and associativity is inherited from matrix multiplication, so ${\mathcal{P}}_n$ is a group once inverses are supplied below.

Taking the adjoint reverses the order and conjugates $i^k$: $(i^k{X}^x{Z}^z{)}^{\ast }=i^{-k}{Z}^z{X}^x=i^{-k}(-1{)}^p{X}^x{Z}^z=P(2p-k,x,z)$, where $p$ represents $x\cdot z$. The product formula then gives $P^{\ast }P=P{P}^{\ast }=I$, so this adjoint is the inverse and completes the group proof.

Scalar operators $i^{k}I$ commute with every matrix and belong to ${\mathcal{P}}_n$. Conversely, if $P(k,x,z)$ commutes with every $X_j$, the product law forces $z_j=0$ for each $j$; commuting with every $Z_j$ forces $x_j=0$ for each $j$. Thus every central element is scalar and the center is exactly $Z_n$.