# Abelian group and faithful representation

Let $$G$$ be an abelian group and for $$i=1...k, \chi_{i}$$ are the characters of the irreducible representations of $$G$$. Denote by $$G'$$ the group of these characters with operation being pointwise multiplication.

Let $$\chi$$ be a character of a representation $$\alpha$$ of $$G$$ s.t. $$\chi = m_{1}\chi_{1}+...m_{k}\chi_{k}$$ where the $$m_{i}'s$$ are nonnegative integers. Let $$S=$$ {$$\chi_{i} | m_{i}>0$$}.

In a paper I'm reading, there is a sentence unclear to me. It goes:

$$\alpha$$ is faithful if and only if $$S$$ generates $$G'$$.

I suppose if $$\alpha$$ is faithful, then the intersection of the kernels of the $$\chi_{i}'s$$ in $$S$$ is trivial, and then...? I'm trying to get closer to the fact that the intersection of the elements of $$G'$$ is trivial...

Any help would be appreciated.

• if the intersection of the kernels is nontrivial the representation is not faithful. On the other hand if it is not faithful the intersection of the kernels is nontrivial, nearly by definition. Commented Nov 25, 2022 at 9:12
• How to connect to $S$ generating $G'$ then? Thank you. Commented Nov 25, 2022 at 9:21
• If $S$ generates the dual group the intersection of the kernels for $\chi\in S$ is equal to the intersection of the kernels of $\chi\in G',$ But the latter is trivial. If $S$ does not generate $G'$ there is $\chi_0\in G$ which is not generated by $S.$ I have impression that there exists an element $g_0\in G$ such that $\chi_0(g_0)\neq$ and $\chi(g_0)=0$ for $\chi\in S.$ Commented Nov 25, 2022 at 10:13
• There are misprints in my comment $\chi_0(g_0)\neq 1$ and $\chi(g_0)=1$ for $\chi\in S.$ Commented Nov 25, 2022 at 11:14

$$(<=)$$ By the fact that for two characters $$\eta$$ and $$\psi$$, we have $$Ker\eta \cap Ker\psi\le Ker\eta\psi$$, we get that $$Ker\chi\le Ker\phi$$, for every irreducible character $$\phi$$ of $$G$$. Hence $$Ker\chi=1$$.
$$(=>)$$ By problem 2.7(b) of Character Theory of Finite Groups (Isaacs, 1976), there exists $$K\le G$$, such that $$=K^\perp$$. We claim that $$K=1$$, which implies that Irr$$(G)=$$. Assume that $$k\in K$$, by the definition of $$K^\perp$$, $$k\in Ker\eta$$, for each $$\eta \in $$, in particular, for each $$\eta \in S$$. So $$k\in Ker\chi=\underset{\eta \in S}{\bigcap}\eta$$. On the other hand, we know that $$Ker\chi=1$$, which means $$K=1$$, as claimed.
Problem 2.7(b): Assume that $$G$$ is abelian. For $$H\le G$$, define $$H^\perp=\{\lambda| H\le Ker\lambda \}$$. Then $$\perp$$ is a bijection from the set of subgroups of $$G$$ to the set of subgroups of Irr$$(G)$$.