Abelian group and faithful representation

Question

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.

Answer 1

$(<=)$ 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 $<S>=K^\perp$. We claim that $K=1$, which implies that Irr$(G)=<S>$. Assume that $k\in K$, by the definition of $K^\perp$, $k\in Ker\eta$, for each $\eta \in <S>$, 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$(G)$.