I am attempting to solve the following problem:

Let $A$ be a finite abelian group, and let $φ:A\to \mathbb{C^\times}$ be a homomorphism that is not the trivial homomorphism. Prove that $\sum_{a\in A} φ(a)=0$.

I know by the structure theorem that $A$ is a direct product of cyclic groups. I have proven it for the special case that A is cyclic, but I need help to prove the general case.

Proof of special case:

Suppose $A$ is cyclic. Say $A=(\mathbb{Z}/n\mathbb{Z}, +)$. Since $φ$ is not trivial, $n\geq2$. Now, we have $φ(a+b)=φ(a)φ(b)$ and $φ(0)=1$. We also have $φ(a)=φ(1)^a$. It follows that $φ(1)\neq 1$, since otherwise $φ$ would be trivial. Now, $0=φ(1)^n-1=[φ(1)^0+φ(1)^1+\cdots+φ(1)^{n-1}][φ(1)-1]$, and therefore $\sum_{a\in A} φ(a)=φ(1)^0+φ(1)^1+\cdots+φ(1)^{n-1}=0,$ as needed. QED.

  • $\begingroup$ Here is a non-optimal solution, but maybe a perspective worth mentioning: $$\rho:=\frac{1}{|G|} \sum_{g \in G} \phi(g)$$ is a projection onto the elements of $\mathbb C$ fixed by the action $\phi:G \to \mathbb C^{\times}$. On one hand, if $v$ is fixed by $\phi(g)$ for all $g$, then $\rho$ certainly acts by identity on $v$, so it is in the image of $\rho$. On the other hand, if $w \in \mathrm{Im}(\rho)$, then $\phi(h) (w)=\frac{1}{|G|}\sum_{g \in G} \phi(h)\phi(g)(v)$ $\endgroup$ Apr 17, 2018 at 13:58
  • $\begingroup$ by linearity, but since $\phi$ is a homomorphism, we know that $\phi(h)\phi(g)=\phi(hg)$, and furthermore, this is just $\frac{1}{|G|}\sum_{hg \in G} \phi(gh)(v)$, which is just $\rho$. $\rho$ is assumed to be nontrivial and linear, so what is the dimension of its image? $\endgroup$ Apr 17, 2018 at 13:59

1 Answer 1


You can also take $b \in A$ such that $\phi (b) \neq 1$ Now $$\sum_{a\in A} \phi (a) = \sum_{a \in A} \phi (a\star b) = \phi (b) \sum_{a \in A} \phi (a)$$ So $$(1-\phi (b) ) \sum_{a\in A} \phi (a) = 0$$ You can conclude from that since $\phi (b) \neq 1$

  • $\begingroup$ (+1) This is a very nice argument that has some geometric interpretation. All the points $\varphi(a)$ are roots of unity, and shifting the sum to $\varphi(a \star b)$ essentially just corresponds to a rotation, which is a shift. But then the symmetry of the set of roots of unity forces the sum to be zero. $\endgroup$
    – user296602
    Apr 17, 2018 at 1:06
  • 2
    $\begingroup$ This proof works even when $A$ is not abelian. $\endgroup$
    – lhf
    Apr 17, 2018 at 1:10
  • $\begingroup$ Wow, I'm surprised that this problem is much simpler than I would have expected. $\endgroup$ Apr 17, 2018 at 1:10
  • $\begingroup$ @lhf Yes but it works. But I think since $\phi$ exists $A$ has to be abelian. $\endgroup$
    – Kroki
    Apr 17, 2018 at 1:11
  • 1
    $\begingroup$ @Youem, not really. Consider $\text{sign}: S_n \to \{-1,1\} \subseteq \mathbb C$. $\endgroup$
    – lhf
    Apr 17, 2018 at 1:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .