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For a subgroup $K$ of a finite abelian group $G$, its annihilator isThese are exactly the linear characters that descend to the quotient group $G/K$. The character group of a finite abelian group has the same size as the group, so $|K^{\perp }|=|G/K|=|G|/|K|$. The annihilator is what abelian hidden-subgroup Fourier sampling measures.

The character group of a topological group that is an abelian group consists of its continuous group homomorphisms to the circle group, with pointwise multiplication. For a finite group, all characters are continuous and their values are roots of unity, so this agrees with the character group of a finite abelian group.

A character of a finite abelian group is a homomorphism into the complex unit circle. Equivalently it is a linear character; finite order forces each character value to be a root of unity. These are the elements of the character group of a finite abelian group and form the orthogonal Fourier basis of functions on the group.

For $a,b\in {\mathbb{Q}}_p^{\times }$, the quadratic Hilbert symbol $(a,b{)}_p$ is one if $b$ is a field norm from ${\mathbb{Q}}_p(\sqrt{a})$, and minus one otherwise; a square $a$ gives the trivial extension and symbol one. It depends only on the two square classes. Equivalently, the conic $z^2=a{x}^2+b{y}^2$ has a nonzero rational point precisely when the symbol is one. Interchanging $a$ and $b$ in the conic gives $(a,b{)}_p=(b,a{)}_p$. The Hilbert norm residue symbol is a group homomorphism in each argument; for nonsquare $a$, the cyclic local norm index makes its group kernel a subgroup of index two. The resulting pairing on square classes is nondegenerate. Alsosince $N(\sqrt{a})=-a$. In the language of local class field theory, $(a,b{)}_p={\mathrm{Art}}_p(b)(\sqrt{a})/\sqrt{a}$; in the Brauer group, it records whether the quaternion cyclic algebra $(a,b)$ splits. These are compatible norm-obstruction descriptions.

An unramified extension of local fields of degree two has norm group consisting precisely of elements with even valuation. Indeed,and the norms on units are surjective. For the latter assertion, the norm on the finite residue fields is surjective. On successive higher principal-unit groups, the congruenceallows one to correct a proposed unit norm one digit at a time: the residue field trace is surjective, and completeness makes the corrections converge. An unramified quadratic field is unique up to isomorphism. Therefore, if $d$ is its defining nonsquare class,

For odd $p$, choose the nonsquare unit $u_0$ from part (i). The polynomial $T^2-u_0$ has irreducible separable reduction, so it defines the unramified quadratic extension. ThusThe trivial class is included in ${\Delta }_2$, since a degree-one extension is unramified. The two units pair trivially, while $(u_0,p{)}_p=-1$. Any class outside ${\Delta }_1$ has odd valuation and consequently pairs nontrivially with $u_0$. This proves directly

For $p=2$, the unit square classes from part (i) give ${\Delta }_1=\{1,-1,5,-5\}$. The element $\omega =(1+\sqrt{5})/2$ satisfies $T^2-T-1$; its reduction $T^2+T+1$ is irreducible and separable over ${\mathbb{F}}_2$. This shows that ${\mathbb{Q}}_2(\sqrt{5})$ is unramified of degree two, givingFor a fully explicit pairing check, use the generators $(-1,5,2)$. We have $(-1,5{)}_2=1$ and $(-1,2{)}_2=1$, since $5=N(1+2i)$ and $2=N(1+i)$ from ${\mathbb{Q}}_2(i)$. But $(-1,-1{)}_2=-1$: a norm equal to $-1$ would be a sum of two squares. If either summand had negative valuation, the sum would have negative valuation (equal negative valuations give valuation $1-2r<0$); if both were integral, their squares could not sum to $-1\equiv 3(\mathrm{mod}4)$. Finally, the unramified norm criterion gives $(5,2{)}_2=-1$. The identity $(a,b{)}_2=(b,a{)}_2$ and $(a,a{)}_2=(a,-1{)}_2$ determine the remaining entries. Writing $(a,b{)}_2=(-1{)}^{B(a,b)}$, the exponent matrix isFor $b=(-1{)}^s{5}^t{2}^r$, annihilating both generators $-1,5$ of ${\Delta }_1$ requires $s=r=0$, so $b\in \{1,5\}$. Annihilating $5$ alone requires $r=0$, so $b$ is a unit class. HenceThese are the Hilbert-symbol annihilators of units and unramified square classes. Annihilators here are taken with respect to this pairing, or equivalently in the character group of a finite abelian group after identifying square classes with their Hilbert-symbol characters.

In this part identify $\mathbb{T}$ with the complex unit circle by $t\mapsto e^{it}$. Write the finite abelian group $G$ additively and let $\hat{G}=\mathrm{Hom}(G,\mathbb{T})$, with pointwise multiplication, be the character group of a finite abelian group of $G$. A linear character has modulus one, since its values have finite order. We first prove the extension and counting facts needed for Fourier analysis on a finite abelian group.

If $H\le G$, $\chi \in \hat{H}$, and $x\notin H$, let $d$ be the least positive integer with $dx\in H$. Choose $z\in \mathbb{T}$ with $z^d=\chi (dx)$ and defineThis is well-defined: two representations differ by $j-j^{\prime }=qd$ and $h-h^{\prime }=-qdx$, whose contributions cancel. It is a group homomorphism and extends $\chi$. Conversely every extension has this form, and the equation for $z$ has exactly $d$ distinct roots. Since $[H+⟨x⟩:H]=d$, successive adjunctions prove that each linear character of $H$ has exactly $[G:H]$ extensions. Taking $H=\{0\}$ gives $\boxed{|\hat{G}|=|G|}$. This is the extension of a character across a cyclic quotient, proved here without a structure theorem. Extending the linear character sending a nonzero element $g$ of order $d$ to $e^{2\pi i/d}$ also proves that linear characters separate points of $G$.

Use the normalized inner product $⟨f,g⟩=|G{|}^{-1}{\sum }_{x\in G}f(x)\overline{g(x)}$. For a nontrivial linear character $\chi$, choose $a$ with $\chi (a)\ne 1$. Translation of the sum gives ${\sum }_x\chi (x)=\chi (a){\sum }_x\chi (x)$, so that sum is zero. Applying this to $\chi \overline{\psi }$ givesThere are $|G|$ such orthonormal vectors in the $|G|$-dimensional space ${\mathbb{C}}^G$, so they form an orthonormal basis. Define the Fourier coefficients by $\hat{f}(\chi )=|G{|}^{-1}{\sum }_{x}f(x)\overline{\chi (x)}$. Expanding in that basis proves Fourier inversion on a finite group and the Parseval identity on a finite group:For normalized convolution $(f\ast g)(x)=|G{|}^{-1}{\sum }_{y}f(y)g(x-y)$, changing the finite summation variables yields $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Translation $f(x)\mapsto f(x+a)$ multiplies its Fourier coefficient by $\chi (a)$. These formulas fix all normalization and conjugation conventions.

The PDF's double-hat identification is canonical Pontryagin duality: define $e:G\to \hat{\hat{G}}$ by $e_x(\chi )=\chi (x)$. This is a group homomorphism and is injective because linear characters separate points. Applying the already proved character-counting result to $\hat{G}$ shows that both groups have size $|G|$, soThe TeX loses one hat. For completeness, the single-dual isomorphism $\hat{G}\cong G$ also holds, but is generally noncanonical. Here is a proof including the needed finite cyclic decomposition. Choose an element $a$ of maximal order $d$. Every element order divides $d$: for two elements, extract their prime-power components by integer multiples: if an element has order $d$ and $p^{\alpha }$ is the exact power of $p$ in $d$, multiplying it by $d/p^{\alpha }$ gives order $p^{\alpha }$. Then combine the components with the larger order for each prime. Components of coprime orders have a sum of product order, because their cyclic subgroups have trivial intersection. This constructs an element of order the least common multiple of the two original orders, which cannot exceed maximal $d$; hence each order divides $d$.

Extend the primitive linear character of $⟨a⟩$ to $\chi :G\to \mathbb{T}$. All its values are $d$th roots because all element orders divide $d$. Define $\rho (g)=ja$ when $\chi (g)=e^{2\pi ij/d}$. This is a group homomorphism fixing $⟨a⟩$, and $g=\rho (g)+(g-\rho (g))$ shows $G=⟨a⟩\oplus \ker \rho$. Induction on group size decomposes $G$ into finite cyclic groups. A cyclic group of order $d_j$ has precisely the linear characters $j\mapsto e^{2\pi imj/d_j}$ and hence a character group of a finite abelian group isomorphic to itself. The dual of a finite direct product of groups is the product of its factor duals, since a linear character is determined by its restrictions. This proves $\hat{G}\cong G$ after choices of cyclic factors and generators, without using an unproved finite-abelian-group structure theorem.

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.

Write the finite abelian group additively. Its group shift operator isThese unitary operators form the regular representation and commute. The representation-theoretic facts we use are that every irreducible representation of a finite abelian group is one-dimensional, there are $|G|$ such characters of a representation, and their character orthogonality gives an orthonormal basis of functions on $G$. Each linear character here is a group homomorphism $\chi :G\to U(1)$, with $|\chi (g)|=1$.

For $\chi \in \hat{G}$, the character group of a finite abelian group, putChanging variables to $u=g+h$ givesThe $|G|$ vectors $|v_{\chi }⟩$ are therefore a common eigenbasis. Using $\chi (g)$ instead of its complex conjugate in their definition reverses every eigenphase; this is merely the quantum Fourier transform sign convention.

Writing $\hat{G}$ for the character group of a finite abelian group, the Bohr set with frequency set $\Gamma \subseteq \hat{G}$ and width $0\le \rho \le 2$ isIts rank is $|\Gamma |$.

For a finite abelian group $G$ and its character group of a finite abelian group $\hat{G}$, the quantum Fourier transform over a finite abelian group isReplacing $\chi (g)$ by its complex conjugate gives the equally common inverse-transform convention.