# Does there exist a general way of finding the size of the stabilizer group $|S|$?

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.

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

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 $$|S| = 2^{n - k}$$