# How to find the order of the error group $E$

The following is taken from "Quantum Error Correction Via Codes Over GF(4)" Calderbank, Rains, Shor, Sloane.

We are told that the group $$E$$ of tensor products $$\pm w_{1} \otimes \dots \otimes w_{n}$$ and $$\pm iw_{1} \otimes \dots \otimes w_{n}$$, where each $$w_{j}$$ is one of $$I, \sigma_{x}, \sigma_{y}, \sigma_{z}$$, describes the possible errors in $$n$$ qubits.

So $$E$$ is a subgroup of the unitary group $$U(2^{n})$$

We want to construct a quantum code from a pair of subgroups of the quantum error group $$E$$. The group $$E$$ has order $$2^{2n + 2}$$.

Would someone be able to explain to me why $$E$$ has order $$2^{2n + 2}$$?

Note that $$2^{2n+2} = 4^{n+1}$$. Now it is easy to solve the combinatorial problem.
There are $$n$$ elements in the tensor product. Each element can be one of four operators $$I, X, Y, Z$$. That gives us $$4^n$$ possibilities.
Finally, there are four phases $$\pm 1, \pm i$$.
Hence $$4^{n+1}$$ operators in the group.