Character group of a finite abelian group ($\hat{G}$)

The character group $\hat{G}$ consists of the group homomorphisms $\gamma :G\to {\mathbb{C}}^{\times }$. Pointwise multiplication makes it a finite abelian group isomorphic to $G$.

A multiplicative character is a group homomorphism from a finite field's multiplicative group of a finite field to the complex unit circle. A generator of the cyclic multiplicative group of a finite field produces a character of every order dividing $q-1$. Extend every such character to zero by value zero, including the trivial multiplicative character, when using cyclotomic indicator formulas.

When $\chi ,\eta ,\chi \eta$ are nontrivial, changing variables in a product of Gauss sums of finite-field characters gives $G(\chi )G(\eta )=J(\chi ,\eta )G(\chi \eta )$, hence $|J|=\sqrt{q}$. If at least one character is nontrivial and their product is trivial, or the other character is trivial, the absolute value is one. Thus all non-main terms in a two-character cyclotomic intersection count are $O(\sqrt{q})$.

Here $\psi$ is a nontrivial additive character of the finite field. For nontrivial $\chi$, substituting $x=ty$ in the squared absolute value givesThe inner sum is $q-1$ at $t=1$ and $-1$ otherwise; multiplicative character orthogonality completes the identity.

Let $H\le G$ be finite abelian and let $\chi :H\to \mathbb{T}$ be a character of a finite abelian group. To adjoin $x\in G$, let $d$ be the least positive integer with $dx\in H$ and choose a complex number $z$ of modulus one with $z^d=\chi (dx)$. Then $\tilde{\chi }(h+jx)=\chi (h)z^j$ is well-defined and extends $\chi$ to $H+⟨x⟩$. There are exactly $d$ choices, corresponding to the $d$ roots. Iterating proves character extension to $G$, proves that $G$ has exactly $|G|$ characters, and proves separation of points by extending a primitive character of the cyclic subgroup generated by a nonzero point. No structure theorem is needed for this argument.

If $H$ is a subgroup of a finite abelian group, $a\notin H$, and $k$ is the least positive integer with $ka\in H$, then each character of a finite abelian group on $H$ has exactly $k$ extensions to $H+⟨a⟩$. The displayed root choice makes the extension well-defined. Starting with the trivial subgroup and adjoining generators proves that the number of characters equals the group order without using a decomposition into cyclic groups.

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 linear character $\chi$ of a finite group $K$,Indeed, if $\chi (h)\ne 1$, the bijection $k\mapsto hk$ gives ${\sum }_k\chi (k)=\chi (h){\sum }_k\chi (k)$, forcing the sum to vanish. The trivial character gives $|K|$ directly. The same argument applies to the restriction of a linear character of a larger group to a subgroup.

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.