$f: G \to \mathbb{C}^*$ is a homomorphism. Show that the sum $\sum f (g) = 0$ or $n$

Question

Let $ \mathbb{C}^*$ be the multiplicative group of non-zero complex numbers. Let $G$ be an abelian group and suppose $f: G \to \mathbb{C}^*$ is a homomorphism. Prove that $\sum_{g \in G} f(g)=n$ or, $\sum_{g \in G} f(g)=0$, where $n =o(G)$

Proof attempt:

The case is evident for the trivial homomorphism; the sum adds up to $n$.

For the second part

We know, the only elements with finite order in the group $ \mathbb{C}^*$ are $1$ and $-1$, with $o(-1)=2$.

Now, the only case when $f(g)$ can take $-1$ as a value is when $n$ is even.

Consider the subgroup $(\{1, -1\}, .) = G'$ of the group $ \mathbb{C}^*$. We have, from the Isomorphism Theorem, $ G/ \ker( f ) \simeq G' $ [since $f$ takes each value from $G'$].

As $o(G')=2$, $o(G/ \ker( f ))=2$, i.e $o(\ker (f))= n/2$. Hence, when summed, the resultant is $0$.

Edit: A foolish assumption has been taken. The finite ordered complex numbers in the said group is of the form $z^n=1$, so I have 'proved' a very restricted case, which is not at all desired.

Answer 1

It is not necessary that $G$ be abelian, to wit:

If

$f(g) = 1, \; \forall g \in G, \tag 1$

then clearly

$\displaystyle \sum_{g \in G} f(g) = n, \tag 2$

since

$o(G) = n; \tag 3$

if

$\exists h \in G, \; f(h) \ne 1, \tag 4$

then since

$hG = G, \tag 5$

we have

$$\begin{align} \sum_{g \in G} f(g) &= \sum_{g \in G} f(hg) \\ &= \sum_{g \in G} f(h)f(g) \\ &= f(h)\sum_{g \in G} f(g); \tag 6 \end{align}$$

with $f(h) \ne 1$ this forces

$\displaystyle \sum_{g \in G} f(g) = 0. \tag 7$

$OE\Delta$.

Answer 2

Here is one novel way using representation theory. The homomorphism $f$ is a (1-dimensional) irreducible representation of a finite group $G$ and $\sum_{g \in G} f(g)$ is the sum of the character $\chi_f$ over $g \in G$, i.e. $\chi_f = f$.

Since $\sum_{g \in G} \chi_f(g) = \lvert G \rvert\langle \chi_f, 1 \rangle$, the sum is zero if and only if $1$ is not a direct summand of $f$. In that case, $f$ is trivial and the sum is $n$.