Binary Pauli quotient isomorphism

By the one-factor actions in [F3] and the factorwise tensor-map rules in [F4], $P(k,x,z)\mathrm{\mid }y\text{〉}=i^k(-1{)}^{z\cdot y}\mathrm{\mid }y+x\text{〉}$. Distinct bit strings give distinct basis vectors by [F2], so the output basis index of the image of |0〉 is x.

Suppose $P(k,x,z)=\lambda P(k^{\mathrm{\prime }},x^{\mathrm{\prime }},z^{\mathrm{\prime }})$ for a nonzero scalar $\lambda$. Applying both sides to $\mathrm{\mid }{0}^n⟩$ gives nonzero multiples of $\mathrm{\mid }x⟩$ and $\mathrm{\mid }{x}^{\mathrm{\prime }}⟩$; distinct bit strings index distinct basis vectors by [F2], so $x=x^{\mathrm{\prime }}$. Comparing the coefficients of this common basis vector gives $i^k=\lambda i^{k^{\mathrm{\prime }}}$. Applying the operators to $\mathrm{\mid }{e}_j⟩$ gives the common output basis vector $\mathrm{\mid }{e}_j+x⟩$ with coefficients $i^k(-1{)}^{z_j}$ and $\lambda i^{k^{\mathrm{\prime }}}(-1{)}^{z_j^{\mathrm{\prime }}}$; their equality and the preceding phase equality imply $z_j=z_j^{\mathrm{\prime }}$ for every $j$. Thus operators equal up to scalar exactly when their binary labels agree; when $\lambda =1$, the same coefficient comparison gives $k=k^{\mathrm{\prime }}$ modulo $4$, so the exact parameter triple is unique.

The parameter uniqueness in step 2.1 makes $\pi$ an unambiguous map on operators. The product formula [F1] makes $\pi$ a homomorphism; its kernel is exactly the zero-label operators $P(k,0,0)=i^{k}I=Z_n$, and every $(x,z)$ is the image of $P(0,x,z)$. The scalar set $Z_n$ is a subgroup because it contains $I$ and is closed under products and inverses; it is central by [F1]. Hence $P\sim Q$ is an equivalence relation: $P^{-1}P=I\in Z_n$, $Q^{-1}P=(P^{-1}Q{)}^{-1}\in Z_n$ whenever $P^{-1}Q\in Z_n$, and $P^{-1}R=(P^{-1}Q)(Q^{-1}R)\in Z_n$ whenever both factors lie in $Z_n$. Its class $[P]$ is the right coset $P{Z}_n$, since $P^{-1}Q\in Z_n$ exactly when $Q=Pz$ for some $z\in Z_n$. If representatives change from $P,Q$ to $Pz,Qw$ with $z,w\in Z_n$, centrality gives $PzQw=PQzw$, so $[P][Q]=[PQ]$ is well-defined. Step 2.1 makes the induced coset map injective, while $P(0,x,z)$ gives surjectivity. It is therefore the asserted quotient isomorphism.

The 2n vectors (e_j,0) and (0,e_j) span V_n by coordinate expansion and are linearly independent because a zero linear combination has each x and z coordinate zero. They form a basis by [F5], so dim V_n=2n.