Question about generators of the group of characters Let $G^*$ be the group of characters of a finite group $G$ and $\phi_1,\cdots , \phi_n$ elements of $G^*$. Is there a criterion to establish if $\phi_1,\cdots , \phi_n$ generate $G^*$?
I would like to say that they generate $G^*$ if and only if $\cap_i \ker(\phi_i)=1$.
One direction is trivial:
Suppose they are generators. Taking $h\neq 1$, there exists $\eta\in G^*$ such that $\eta(h)\neq 1$. However $\eta$ is written as a combination of $\phi_i$ and so if by contradiction  $h\in \cap_i \ker(\phi_i)$, then $\eta(h)=1$.
Maybe does this work for $G$ abelian?
Any suggestions are appreciate
 A: $\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\C}{\mathbb{C}}$This is a rephrasing of OP's solution.
Let $G$ be a finite abelian group, multiplicatively written. We will use (standard) duality arguments.
Consider the bilinear map $G^{*} \times G \to \C^{\star}$ given by $(\chi, g) \mapsto \chi(g)$. For $B \subseteq G$, let $B^{\perp} = \Set{ \chi \in G^{*} : \chi(g) = 1}$, a subgroup of $G^{*}$, and similarly for $B \subseteq G^{*}$.
Note that $B \subseteq B^{\perp\perp}$, and $B \subseteq C$ implies $C^{\perp} \subseteq B^{\perp}$, and thus $B^{\perp\perp} \subseteq C^{\perp\perp}$.
Moreover, if $H \le G$, one has that each character of $H$ can be extended in $\Size{G : H}$ ways to characters of $G$ (see e.g. Theorem 3.4 here). In particular, if one starts with the trivial character of $H$, then this can be extended to all characters of $G/H$ (slight abuse of language here), so that $H^{\perp\perp} = H$. The same argument applies to $H \le G^{*}$.
Let $B \subseteq G^{*}$, and let $\Span{B}$ be the subgroup it generates. From $B \subseteq \Span{B}$ we get $B^{\perp\perp} \subseteq \Span{B}^{\perp\perp} = \Span{B}$. Now $B \subseteq B^{\perp\perp} \le G^{*}$ implies $B^{\perp\perp} = \Span{B}$, since $\Span{B}$ is the smallest subgroup of $G^{*}$ containing $B$.
Finally, if the intersection of all the kernels of the elements of $B \subseteq G^{*}$ is $\Set{1}$, that is, $B^{\perp} = \Set{1}$, then
$$
\Span{B}
=
B^{\perp\perp}
=
\Set{1}^{\perp}
=
G^{*}.
$$
A: I answer to my question because maybe I found a solution.
If $G$ is not abelian, then the sentence is not true, as explained by Captain Lama.
Let us suppose $G$ be a finite abelian group.
I want to prove that a set of characters $\phi_1,\cdots \phi_n$ generate $G^*$ if and only if $\cap_i \ker(\phi_i)=1$
The implication $\Longrightarrow$ is already explained in my question and it works because $G$ is a finite abelian group.
Thus consider $\Longleftarrow$
By contradiction $K:= \langle \phi_i \rangle \neq G^*$.
$Lemma$ 1 : (it’s the equivalence of the consequence of Hahn-Banach theorem in the case of Characters of a Group instead of vector spaces)
Let $G$ be a finite abelian  group and $H$ be a subgroup of $G$. Then there exists a non trivial character $\chi \in G^*$ such that $\chi_H$ is trivial.
$Lemma$ 2 : (it’s the Equivalent result of when a vector space is reflexive)
Let $ G$ be a finite abelian group. The canonical injection $j\colon G \to G^{**}$ sending $g $ to $j(g)(\phi):=\phi(g)$ is an isomorphism.
By Lemma 1, take a non trivial character $F\in G^{**}$ of $G^*$ such that $F$ is trivial on $K$. By lemma 2, the canonical injection $j$ is surjective and so $F=j(h)$ for some $h\neq 0$.
Then $\phi_i(h)=j(h)(\phi_i)=F(\phi_i)=1$ for each $i$, and so $h\in \cap_i \ker(\phi_i)$.
This is a contradiction.
