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 $\left|{G}\right|$.

Lemma
Let $H \le G$ be a subgroup, and $\chi : H \to \C^\times$ a character.

Let $a \in G \backslash H$ and $n$ be the indicator of $a$ in $H$.

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

Proof
We induct on $\left[{G : H}\right]$, the index of $H$ in $G$.

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

From Enlarging a Subgroup, 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 $\left|{K}\right| = n \left|{H}\right|$.

If $\tilde \chi$ is a character extending $\chi$, then we must have:


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

and also


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

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


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

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

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:

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 $G$.