How many elements are present in $N(S)$ where $S$ is a stabilizer group?

Question

Question

Suppose I have $n$ qubits and $n-k$ stabilizer generators. Let the set of stabilizers be $S$. I then have $k$ logical qubits. Define $N(S)$ to be

$$N(S) = \{p \in P_n\ |\ sp = ps\ \forall s\in S\},$$

where $P_n$ is an $n$-qubit Pauli.

How many generators do I need for the group $N(S)$ and the quotient group $N(S)/S$?

My attempt:

We have $4^n$ elements in $P_n$ (ignoring phases). This is because for each qubit, I can have $I, X, Y$ or $Z$. I also have $2^{n-k}$ stabilizers. That leaves $2^{2n} - 2^{n-k}$ Paulis that I need to split between $N(S)$ and elements that are not in $N(S)$.

Answer 1

Note that the Pauli group up to phases forms a binary vector space $\mathbb{F}_2^{2n}$ where addition corresponds to multiplication of Paulis (this is the usual mapping used to write down check matrices). This perspective is especially useful for counting problems as the one presented.

Since we have have $n-k$ independent generators $g_i$, it is sufficient to consider the system of equations given by $p g_i = g_i p$. This corresponds to a linear system of equations with $n-k$ constraints over $\mathbb{F}_2^{2n}$. Hence, its solution space in $\mathbb{F}_2^{2n}$ has dimension $2n - (n-k) = n+k$ and thus $2^{n+k}$ elements. Re-introducing phases, this means that $|N(S)| = 4\times 2^{n+k}$. Now, $S\subset N(S)$ and $|N(S)/S|=2^{n+k-(n-k)} = 2^{2k}$ which implies that you need $2k$ generators.