So I know that, for a stabilizer code, the stabilizer group $S$ has $n-k$ commuting generators.

Is there a general way of knowing what the order of the full group of $S$ is, aside from writing out all the possible combinations of its generators $g_{1}, \dots, g_{n-k}$ which give rise to different/unique stabilizers?

My intuition says no, because $|S|$ is dependent on the stabilizer generators, which are different for each code. For example, for one code $g_{1}g_{2}=g_{3}g_{4}$, and so, we cannot count this stabilizer twice in $S$. However, for another code $g_{1}g_{2} \neq g_{3}g_{4}$, giving two new stabilizers to $S$. However, I would be grateful if somebody could confirm/debunk my suspicions.

  • 2
    $\begingroup$ If I understand correctly, the cardinality of a full stabilizer group is equal to $2^{n-k}$ $\endgroup$
    – Yunzhe
    Commented Jun 7 at 10:54
  • $\begingroup$ Yes, I think that I was missing the fact that each unique multiplicative combination of stabilizer generators will give rise to unique stabilizer elements, $\endgroup$
    – am567
    Commented Jun 7 at 11:05

1 Answer 1


For a $[[n, k, d]]$ stabilizer code, the $n - k$ stabilizer generators are not only commuting, but they are also independent.

Taking into account that $g^2 = 1$ for each generator $g$, we have the cardinality $$|S| = 2^{n - k}$$


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