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 $|G|$.

Lemma
Let $H \leq G$ 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 $[G:H]$ distinct characters on $G$.

Proof
We induct on $[G:H]$, the index of $H$ in $G$.

If $[G:H] = 1$, then $H = G$ by Langrange's theorem, so the result is trivial.

Now suppose that $[G:H] > 1$, 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 $K = \langle H, a\rangle$.

By Enlarging a Subgroup we know that each element of $K$ has a unique representation in the form $x a^k$ with $x \in H$ and $0 \leq k \leq n-1$, and $|K| = n|H|$.

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


 * $\tilde\chi(x a^n) = \tilde\chi(x) \tilde \chi(a)^n \qquad (1)$

and also


 * $\tilde\chi(x a^n) = \tilde\chi(x) \tilde \chi(a^n) \qquad (2)$

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


 * $\tilde \chi(a)^n = \chi(a^n)$

That is $\tilde\chi(a)$ is an $n^\text{th}$ root of $\chi(a^n)$, so by nth roots of a complex number are distinct there are $n$ distinct possibilities.

We choose one of these possibilities and define $\tilde\chi(x a^k) = \chi(x) \tilde\chi(a)^k$, for $x a^k \in K$.

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

Now by Lagrange's theorem, $n = |K| / |H| = [K : H]$

Also by Lagrange's Theorem, $[G : K] < [G:H]$, since $H \subsetneqq K$.

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

Therefore by the tower law for subgroups, $\chi$ extends to $[K : H][G : K] = [G : H]$ characters on $G$.

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

Therefore there is only one character $\chi : \{e\} \to \C^\times : e \mapsto 1$ on the Trivial Subgroup.

Moreover the trivial subgroup has index $[G : \{e\}] = |G|$ (Langrange's theorem), so by the lemma there are $|G|$ distinct characters on $G$.