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

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.

• What's the order of $i$ then? – the_fox Jan 4 at 3:11
• Does G have to have finite order? – Joel Pereira Jan 4 at 3:39
• @JoelPereira:. isn't that implied by the statements $\sum f(g) = n$ and $n = o(G)$? – Robert Lewis Jan 4 at 3:45
• @the_fox :( back to square one. – Subhasis Biswas Jan 4 at 11:41
• @RobertLewis no it's not implied. If G = the multiplicative group of $\mathbb{R}^+$, we can still form the sum. In that case the sum would diverge. – Joel Pereira Jan 4 at 15:33

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

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

$$OE\Delta$$.

• @Shaun: nice edit, thanks! – Robert Lewis Jan 4 at 6:23
• Amazing. Just amazing. – Subhasis Biswas Jan 4 at 11:40
• Interesting. I wonder if they just assumed abelian so that students could use the fundamental theorem. It's very easy to prove for cyclic groups. – Cameron Williams Jan 4 at 12:03
• @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers! – Robert Lewis Jan 5 at 0:41

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 $$n$$.

• Can you please verify mine? – Subhasis Biswas Jan 4 at 6:00
• the_fox in one of the earlier comments has already pointed out that a mistake in your proof is assuming the only elements with finite order in $\mathbb{C}^*$ are $\pm 1$, when in fact any complex $z$ such that $z^\ell = 1$ for a nonzero integer $\ell$, i.e. a root of unity, has finite order at most $\ell$. – Riley Jan 4 at 7:01
• $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it? – Subhasis Biswas Jan 4 at 8:27
• I'm not sure if I can adapt your approach. I might give this a go later myself, but you might be able to adapt it by first using the structure theorem for finitely generated abelian groups to first decompose $G$ into finite cyclic groups $\mathbb{Z}_{m}$. On the direct summand $\mathbb{Z}_{m}$, if $g_i$ is a generator, so that $g_i^m = 1$, then if we let $z = f(g_i)$ then $z^m = 1$ and $1 + z + \cdots + z^{m-1} = 0$ if $z \neq 1$. – Riley Jan 4 at 10:37
• I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :) – Subhasis Biswas Jan 4 at 11:40