Fourier analysis on finite abelian groups

Question

Can someone help me show if $f$ is a character of a finite abelian group then for all $a\in G$,

$$\sum_{[f]}f(a)\stackrel{}{=} \begin{cases} |G| & \text{if $ a$ is the identity} \\ 0 & \text{otherwise} \end{cases}$$

Where the sum runs over all characters of $G$,

I was thinking of using the fact that the characters themselves form an abelian group under multiplication, so that I have$$f_k(a)\sum_{[f]}f(a)=\sum_{[f]}f(a)f_k(a)=\sum_{[f]}f(a)$$

So that, $$(1-f_k(a))\sum_{[f]}f(a)=0$$ But I would need $\exists k|f_k(a)\ne 1, \forall a \in G, a\ne e$

In order to divided both sides by $(1-f_k(a))$, which I can't seem to get either.

Answer 1

The proof (which is very standard) can for instance be found in $\S 4.3$ of Appendix B of these notes.

If you take a look, you'll see that you're on the right track, but perhaps you've been stymied by a quantifier error. You want to show your identity for each fixed nonidentity element $a \in G$. So you don't need to find a character which is nontrivial on all nonidentity elements (which is not always possible; consider e.g. $G = Z/2 \times Z/2)$. You just need that each nonidentity element is not killed by at least one character on $G$. There is still something to show here, and the notes explain this point carefully. (It follows from something called the Character Extension Lemma...)

Answer 2

This is the so called column orthogonality relation, applied to the conjugacy classes of $a$ and $1$.

The proof can be found in any book on representation theory. One first shows that the irreducible representations are a basis for the vector space of class functions, then one expresses one of the standard basis class functions (one that is 1 on one class and 0 on all other classes) as a sum of irreducible characters. One can compute the coefficients explicitly in this sum and then fill in $a$ in the relation that you obtain (noting that $|g^G|=1$, since $G$ is abelian).

Answer 3

Your proof strategy is correct. To prove that there is a character with $\chi(a)\neq 1$ look at Serre's A course in arithmetic:

Proposition 1.1.: Let $H$ be a subgroup of $G$ (finite), then every character $\chi$ on $H$ extends to a character on $G$.

Proof (Sketch): By induction on $(G:H)$. Let $H$ be a non-trivial subgroup, $x\notin H$, and $n$ minimal such that $x^n\in H$. If $\chi(x^n)=t$ choose any $w\in\mathbb{C}^{*}$ with $w^n=t$ and define $\chi'(x)=w$. This gives an extension of $\chi$ to the subgroup of $G$ generated by $H$ and $x$.