1
$\begingroup$

How does one show that for any non-trivial abelian group, there exists some non-trivial character?

After looking up for a while, the best solution I could find was Pontrayagin duality, which seems like too heavy a machinery for this problem.

  • 0
    See [this question](http://math.stackexchange.com/questions/1681017/is-there-a-non-trivial-character-on-any-locally-compact-abelian-group). Is it true that you allow arbitrary abelian groups? – 2017-01-31
  • 0
    @DietrichBurde I have looked at that, I was hoping for an easier solution. – 2017-02-01

2 Answers 2

3

This follows from the fact that $GL_1(\mathbb{C})$ is a divisible and hence injective abelian group. So given any abelian group $A$, a subgroup $B\subseteq A$, and a homomorphism $f:B\to GL_1(\mathbb{C})$, $f$ can be extended to a homomorphism $A\to GL_1(\mathbb{C})$. Now if $A$ is any nontrivial abelian group, let $B$ be a nontrivial cyclic subgroup of $A$. It is easy to explicitly construct a nontrivial homomorphism $B\to GL_1(\mathbb{C})$, which then extends to $A$.

(I doubt you will find any simpler proof than this for arbitrary abelian groups--I'm pretty sure this is impossible to prove without using the axiom of choice, for instance.)

  • 0
    Baer's criterion needs Zorn's lemma, right? – 2017-01-31
  • 0
    Right, that's where the axiom of choice is used in this argument. – 2017-01-31
  • 0
    Nice. Thanks a lot. By the way, a natural question that cropped up in my mind is, does Pontryagin duality require the axiom of choice? I don't know it's proof, so I can't comment myself. – 2017-02-01
  • 0
    @MathManiac: Yes, Pontryagin duality uses the axiom of choice. – 2017-02-01
1

For arbitrary abelian groups something non-trivial has to be used, it seems. For finite abelian groups, though, we have a very elementary Lemma:

Lemma: Let $G$ be a finite abelian group and $a,b$ distinct elements in $G$. Then there exists a character $\chi$ of $G$ such that $\chi(a)\neq \chi(b)$.

For a proof see here, Lemma $1.3.33$. So for $n=|G|\ge 2$ we see that there exists a non-trivial character of $G$.

  • 0
    I think the OP is mainly concerned with the less easy case of non finitely generated groups. – 2017-01-31
  • 0
    I see. Then we need more. Actually, Noam Elkies says we do need Pontryagin duality, see [here](http://math.stackexchange.com/questions/1681017/is-there-a-non-trivial-character-on-any-locally-compact-abelian-group). – 2017-01-31