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For fixed odd modulus, the Jacobi symbol is a homomorphism on the unit group

Statement

Fix an odd positive integer n. The assignment χn([a]n)=(an) is a group homomorphism (Z/n)×{±1}.

Here {±1} is the two-element multiplicative group, except that the image is the one-element subgroup {1} when the character is trivial.

Facts & Assumptions

Given: An odd positive integer n and unit classes [a]n,[b]n(Z/n)×.

[L1]

The Jacobi symbol belongs to {1,0,1}, depends only on a(modn), and is zero exactly when gcd(a,n)>1 (The Jacobi symbol is well defined on numerator residue classes).

[L2]

For odd positive n, (abn)=(an)(bn) (The Jacobi symbol is multiplicative in numerator and denominator).

[L3]

The unit group (Z/n)× consists of the invertible residue classes modulo n under multiplication (The unit group (Z/n)× and Euler's totient φ(n)=(Z/n)× for n1).

[L4]

The class [a]n is a unit if and only if gcd(a,n)=1 (For n1, [a]n is a unit if and only if gcd(a,n)=1).

[L5]

A group homomorphism f:GH is a function satisfying f(xy)=f(x)f(y) for all x,yG (Monoid homomorphism and group homomorphism).

Proof

technique · direct
1.1

By [L1], the value (an) depends only on the residue class. By [L3] and [L4], a unit class has gcd(a,n)=1, so [L1] rules out the value zero; hence χn is a well-defined function from (Z/n)× to {±1}.

L1L3L4
2.1

For unit classes [a]n and [b]n, [L2] gives χn([a]n[b]n)=χn([ab]n)=χn([a]n)χn([b]n), which is the condition in [L5]. Thus χn is a group homomorphism, including when n=1 and the unit group has one element.

step 1.1L2L5

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 82 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources