Graph state is the unique joint stabilizer ray

The vector $\mathrm{\mid }+⟩=2^{-1\mathrm{/}2}(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)$ has norm one by [F8], and $\mathrm{\mid }+{⟩}^{\otimes V}$ has norm one by the product inner product [F6]. Every edge gate in [F2] is unitary by [F5], so their finite product preserves norm and $\mathrm{\mid }G⟩$ is a unit vector. When $V=\mathrm{\varnothing }$, the empty tensor and edge product give the unit vector $1\in \mathbb{C}$ directly.

If $n>0$, every product of the commuting involutions in [F1] is indexed by a subset of the $n$ generators. Its binary $X$-support is the indicator of that subset, so the independent labels in [F1] make the $2^n$ products distinct. Every product fixes the nonzero vector $\mathrm{\mid }G⟩$ from step 1.1, so the generated group cannot contain $-I$. It is a stabilizer subgroup by [F7] and has rank $n$.

For $n>0$, apply [F3] to the subgroup in step 2.1 to identify its group average with the common $+1$ projector. Its rank is $2^{n-n}=1$ by [F4]. The graph state fixes this line by [F1] and is unit by step 1.1, so the orthogonal projector onto the line is $\mathrm{\mid }G⟩⟨G\mathrm{\mid }$ by [F9]. If $n=0$, the empty generator family generates $\{I\}$, whose average is the identity on $\mathbb{C}$; this is also $\mathrm{\mid }1⟩⟨1\mathrm{\mid }$.