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Measuring the coset state $|g_0+K⟩$ in the group shift operator eigenbasis samples uniformly from the annihilator of a subgroup of a finite abelian group $K^{\perp }$. Indeed,so the character-sum cancellation lemma gives probability $|K|/|G|$ on $K^{\perp }$ and zero elsewhere. The modulus of $\chi (g_0)$ is one, so the offset has no effect on the distribution.

Write $x=g^{-1}h$ and $S(x)={\sum }_{\chi \in \hat{G}}\chi (x)$, where $\hat{G}$ is the character group of a finite abelian group. The Fundamental theorem of finitely generated abelian groups writes the finite group $G$ as a product of cyclic groups. A character of a cyclic group factor of order $r$ is determined by an arbitrary $r$th root of unity, so $|\hat{G}|=|G|$. This description also shows that the characters separate points: if $x\ne 1$, a nonzero coordinate of $x$ is detected by a character $\psi$ with $\psi (x)\ne 1$.

Multiplication by $\psi$ permutes $\hat{G}$, soFor $x\ne 1$ this forces $S(x)=0$. For $x=1$ every summand is one. Since $\chi (g{)}^{-1}\chi (h)=\chi (g^{-1}h)$, the character-sum cancellation lemma givesAll character values are roots of unity, so inversion here is also complex conjugation. This is the finite-abelian version of character orthogonality.

Measure the coset state in the common eigenbasis $\{|v_{\chi }⟩\}$ of the group shift operators. Its overlap isThe character-sum cancellation lemma makes this sum $|K|$ when $\chi$ is trivial on $K$, and zero otherwise. Hence, with $K^{\perp }$ the annihilator of a subgroup of a finite abelian group,Since $|K^{\perp }|=|G|/|K|$, this is the uniform distribution on a finite set $K^{\perp }$. The factor $\chi (g_0)$ has modulus one, so the distribution is independent of $g_0$. This is abelian hidden-subgroup Fourier sampling.