The Exercise

A stabilizer group $S$ on $n$ qubits is given by a list of generators $g_1, \dots, g_l$, operators on the $n$ qubits that are tensor products of Pauli matrices up to a phase, chosen so that $-I$ is not in $S$. Each generator is recorded as a row of $2n$ bits, its check row, which says on which qubits it acts as $X$, $Y$ or $Z$.

The Pauli Group

The Pauli group $G_n$ on $n$ qubits is the group of operators $$g = i^s\, \sigma_{a_1} \otimes \sigma_{a_2} \otimes \cdots \otimes \sigma_{a_n}, \qquad s \in \{0, 1, 2, 3\}, \quad \sigma_{a_k} \in \{I, X, Y, Z\},$$(1)^✦ where $i^s$ is the phase of $g$ and the tensor product is its Pauli string. The element $-I$ of $G_n$ is $i^2\, I \otimes \cdots \otimes I$.^✦

Check Rows

The check bits $(x, z)$ of a Pauli matrix, its $X$ bit and its $Z$ bit, are elements of $\mathbb{Z}_2 = \{0, 1\}$, the integers modulo $2$, given by $$I \mapsto (0, 0), \qquad X \mapsto (1, 0), \qquad Y \mapsto (1, 1), \qquad Z \mapsto (0, 1).$$(2)^✦ The check row $r(g)$ of an element $g$ of $G_n$ as in (1) is the row of $2n$ bits $$r(g) = (x_1, \dots, x_n \mid z_1, \dots, z_n) \in \mathbb{Z}_2^{2n},$$(3)^✦ where $(x_k, z_k)$ are the check bits of the $k$th factor $\sigma_{a_k}$ of its Pauli string. Rows are added bit by bit, modulo $2$.

Independent Generators

The group $\langle g_1, \dots, g_l \rangle$ generated by elements $g_1, \dots, g_l$ of a group $G$ is the smallest subgroup of $G$ that contains all of them. Elements $g_1, \dots, g_l$ of $G$ are independent generators when leaving out any one of them generates a smaller group: for every $i$, $$\langle g_1, \dots, g_{i-1}, g_{i+1}, \dots, g_l \rangle \neq \langle g_1, \dots, g_l \rangle .$$(4)

The Claim

What Is Checked

The claim is stated over the Pauli group $G_n$ of (1) with its element $-I$, the check row (3) built from the check bits (2), and independence of generators (4), which is defined for elements of any group and taken here in $G_n$.

The check row of an element of $G_n$ depends on its Pauli string alone, not on its phase $i^s$, and the check row of $-I$ is the zero row.