# A question on the structure of the Clifford group

Let $$P_n$$ be the Pauli group on $$n$$ qubits defined by $$P_n=\{i^k \sigma_1...\sigma_n: k=0, 1, 2 \; \mathrm{or}\;3, \sigma_j\in \{I, X, Y, Z\}\}$$, where $$I, X, Y, Z$$ are Pauli matrices. The Clifford group $$C_n$$ is defined as the normalizer of $$P_n$$ in the unitary group as follows: $$C_n=\{U\in U(2^n):UP_n U^\dagger=P_n\}$$. Is the following true?

$$\{U\in C_n:U\sigma=\pm\sigma U \;\forall\sigma\in P_n\}=P_n$$.

Thank you.

Yes, if a Clifford operation $$C$$ conjugates every Pauli product $$P$$ into $$\pm P$$, then $$C$$ is a Pauli product.
First, show that $$PAP^\dagger = \pm A$$ for any Pauli products $$A$$ and $$P$$. In other words, paulis-implies-only-sign-change. This just comes down to all pairs of Pauli products commuting or anti-commuting.
Second, show only-sign-change-implies-paulis. Take a promised only-sign-change operation $$C$$ and compute $$C B C^\dagger$$ versus the generators $$B \in \{X_1, Z_1, X_2, Z_2, \dots\}$$. The sign changes you get for $$X_1$$ and $$Z_1$$ reveal which of the four pauli operators (IXYZ) is being to qubit 1 by $$C$$, and same for each other qubit index $$k$$. Together they give you a reconstructed Pauli product, which you know is equal to $$C$$ because this operation and $$C$$ do the same thing when conjugating any generator $$B$$.