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.

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    $\begingroup$ What's the order of $i$ then? $\endgroup$ – the_fox Jan 4 at 3:11
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    $\begingroup$ Does G have to have finite order? $\endgroup$ – Joel Pereira Jan 4 at 3:39
  • $\begingroup$ @JoelPereira:. isn't that implied by the statements $\sum f(g) = n$ and $n = o(G)$? $\endgroup$ – Robert Lewis Jan 4 at 3:45
  • $\begingroup$ @the_fox :( back to square one. $\endgroup$ – Subhasis Biswas Jan 4 at 11:41
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    $\begingroup$ @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. $\endgroup$ – Joel Pereira Jan 4 at 15:33

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


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

then clearly

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


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


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

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

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


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    $\begingroup$ @Shaun: nice edit, thanks! $\endgroup$ – Robert Lewis Jan 4 at 6:23
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    $\begingroup$ Amazing. Just amazing. $\endgroup$ – Subhasis Biswas Jan 4 at 11:40
  • $\begingroup$ 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. $\endgroup$ – Cameron Williams Jan 4 at 12:03
  • $\begingroup$ @SubhasisBiswas: thank you for your kind words. If you really like my answer, you might consider "accepting" it. Cheers! $\endgroup$ – 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 $n$.

  • $\begingroup$ Can you please verify mine? $\endgroup$ – Subhasis Biswas Jan 4 at 6:00
  • $\begingroup$ 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 $\ell$. $\endgroup$ – Riley Jan 4 at 7:01
  • $\begingroup$ $z^n=1$ does form a group. Now, can we somehow follow my approach to prove it? $\endgroup$ – Subhasis Biswas Jan 4 at 8:27
  • $\begingroup$ 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 $z \neq 1$. $\endgroup$ – Riley Jan 4 at 10:37
  • $\begingroup$ I was thinking exactly along this line. Now, conversion of this into isomorphism theorem would be really nice. :) $\endgroup$ – Subhasis Biswas Jan 4 at 11:40

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