Number of Characters on Finite Abelian Group

Theorem
Let $G$ be a finite abelian group.

Then the number of characters $G \to \C^\times$ is $\left\vert{G}\right\vert$.

Lemma
Let $H \le G$ be a subgroup.

Let $\chi: H \to \C^\times$ be a character on $G$.

Let $a \in G \mathrel \backslash H$ where $\backslash$ denotes divisibility.

Then:
 * $\chi$ extends to $\left[{G : H}\right]$ distinct characters on $G$

where $\left[{G : H}\right]$ denotes the index of $H$ in $G$.

Proof
The proof proceeds by stroing induction on $\left[{G : H}\right]$.

If $\left[{G : H}\right] = 1$, then $H = G$ by Langrange's theorem.

Hence (trivially) the result.

Suppose that $\left[{G : H}\right] > 1$, and the result holds for all subgroups of smaller index in $G$.

Let $a \notin H$ and let $n$ be the indicator of $a$ in $H$.

Let $K = \left\langle{H, a}\right \rangle$ denote the subgroup generated by $H$ and $a$.

From Subgroup Generated by Subgroup and Element, each element of $K$ has a unique representation in the form $x a^k$ with:
 * $x \in H$ and $0 \le k \le n-1$

and:
 * $\left|{K}\right| = n \left|{H}\right|$

Let $\tilde \chi$ be a character extending $\chi$.

Then:


 * $(1): \quad \tilde \chi \left({x a^n}\right) = \tilde \chi \left({x}\right) \tilde \chi \left({a}\right)^n$

and:


 * $(2): \quad \tilde \chi \left({x a^n}\right) = \tilde \chi \left({x}\right) \tilde \chi \left({a^n}\right)$

Since $x, a^n \in H$, $\tilde \chi$ and $\chi$ agree on these values, so equating $(1)$ and $(2)$ we obtain:


 * $\tilde \chi \left({a}\right)^n = \tilde \chi \left({a^n}\right)$

That is:
 * $\tilde \chi \left({a}\right)$ is an $n$th root of $\tilde \chi \left({a^n}\right)$.

So by nth roots of a complex number are distinct there are $n$ distinct possibilities.

We choose one of these possibilities and define:
 * $\tilde \chi \left({x a^k}\right) = \chi \left({x}\right) \tilde \chi \left({a}\right)^k$

for $x a^k \in K$.

We check that $\tilde \chi$ is multiplicative:

For $x,y\in H$,

By Lagrange's theorem:
 * $n = \dfrac {\left|{K}\right|} {\left|{H}\right|} = \left[{K : H}\right]$

Also by Lagrange's Theorem:
 * $\left[{G : K}\right] < \left[{G : H}\right]$

since $H \subsetneq K$.

So by the induction hypothesis, each $\tilde \chi$ extends to $\left[{G : K}\right]$ characters on $G$.

Therefore by the Tower Law for Subgroups, $\chi$ extends to $\left[{K : H}\right] \left[{G : K}\right] = \left[{G : H}\right]$ characters on $G$.

Since a character is a homomorphism, it must preserve the identity.

Therefore there is only one character $\chi : \left\{{e}\right\} \to \C^\times : e \mapsto 1$ on the trivial subgroup.

Moreover, by Langrange's theorem, the trivial subgroup has index $\left[{G : \left\{{e}\right\}}\right] = \left|{G}\right|$.

So by the lemma there are $\left|{G}\right|$ distinct characters on $G$.