Number of Characters on Finite Abelian Group

This page has been identified as a candidate for refactoring of basic complexity. In particular: Separate the lemma out. Until this has been finished, please leave {{Refactor}} in the code. New contributors: Refactoring is a task which is expected to be undertaken by experienced editors only. Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Refactor}} from the code.

This article needs to be linked to other articles. In particular: one or two places You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.

Theorem

Let $G$ be a finite abelian group.

Then the number of characters $G \to \C^\times$ is $\order G$.

Proof

Lemma

Let $H \le G$ be a subgroup.

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

Let $a \in G \divides H$ where $\divides$ denotes divisibility.

Then:

$\chi$ extends to $\index G H$ distinct characters on $G$

where $\index G H$ denotes the index of $H$ in $G$.

Proof

The proof proceeds by stroing induction on $\index G H$.

If $\index G H = 1$, then $H = G$ by Langrange's theorem.

Hence (trivially) the result.

Suppose that $\index G H > 1$, 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 $K = \gen {H, a}$ 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:

$\order K = n \order H$

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

Then:

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

and:

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

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

$\map {\tilde \chi} a^n = \map {\tilde \chi} {a^n}$

That is:

$\map {\tilde \chi} a$ is an $n$th root of $\map {\tilde \chi} {a^n}$.

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

We choose one of these possibilities and define:

$\map {\tilde \chi} {x a^k} = \map \chi x \map {\tilde \chi} a^k$

for $x a^k \in K$.

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

For $x, y \in H$:

\(\ds \map {\tilde \chi} {x a^k y a^l}\)

\(=\)

\(\ds \map {\tilde \chi} {x y a^{k + l} }\)

\(\ds \)

\(=\)

\(\ds \map \chi {x y} \map {\tilde \chi} a^{k + l}\)

\(\ds \)

\(=\)

\(\ds \paren {\map \chi x \map {\tilde \chi} a^k} \paren {\map \chi y \map {\tilde \chi} a^l}\)

\(\ds \)

\(=\)

\(\ds \map {\tilde \chi} {x a^k} \map {\tilde \chi} {y a^l}\)

By Lagrange's theorem:

$n = \dfrac {\order K} {\order H} = \index K H$

Also by Lagrange's Theorem:

$\index G K < \index G H$

since $H \subsetneq K$.

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

Therefore by the Tower Law for Subgroups, $\chi$ extends to $\index K H \index G K = \index G H$ characters on $G$.

$\Box$

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

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

Moreover, by Langrange's theorem, the trivial subgroup has index $\index G {\set e} = \order G$.

So by the lemma there are $\order G$ distinct characters on $G$.

$\blacksquare$