The Exercise

Let $S$ be a subgroup of the Pauli group on $n$ qubits, and suppose that $S$ does not contain $-I$, minus the identity on all $n$ qubits. The exercise asks to show that $S$ then contains neither $iI$ nor $-iI$.

The Pauli Group

The Pauli group $G_n$ on $n$ qubits is the group of operators $$i^s\, P_1 \otimes P_2 \otimes \cdots \otimes P_n ,$$(1)^✦ where the phase power $s$ is an integer modulo $4$ that fixes the overall factor $i^s$, and the Pauli string $P_1, \dots, P_n$ assigns to each qubit one of the four Paulis $I$, $X$, $Y$, $Z$.

The element $-I$ of $G_n$ is the identity string $I \otimes \cdots \otimes I$ with phase power $s = 2$, which is minus the identity $I$ on all $n$ qubits: $$-I = i^2\, I \otimes \cdots \otimes I .$$(2)^✦ The element $iI$ of $G_n$ is the identity string with phase power $s = 1$, and the element $-iI$ the identity string with phase power $s = 3$: $$iI = i\, I \otimes \cdots \otimes I ,$$(3)^✦ $$-iI = i^3\, I \otimes \cdots \otimes I .$$(4)^✦

The Claim

What Is Checked

The claim is stated over the Pauli group $G_n$ of (1), for any number $n$ of qubits, over an arbitrary subgroup $S$ of it, and over its elements $-I$, $iI$ and $-iI$ of (2), (3) and (4).