If $\chi(a)=1$ for all $\chi\in\hat G$ then $a=0$.

Question

Let $G$ be finite abelian group and $\hat G$ be its character group.

I need hint proving that if $a\in G$ and $\chi(a)=1$ for all $\chi\in\hat G$ then $a=0$ (the identity element).

I can prove it for cyclic groups but not in general.

Answer 1

If you can prove it for the special case of cyclic groups, then you can reduce the general case to the special one.

Let us say you want to prove that if $a \ne 0$, then there is a character $\chi$ of $G$ such that $\chi(a) \ne 1$.

Show that there is a subgroup $N$ of $G$ such that $a \notin N$ (that is, $a N \ne 0$ in $G/N$) and $G/N$ is cyclic.

Then take a character $\psi$ of $G/N$ such that $\psi(aN) \ne 1$, and lift $\psi$ to the character $\chi$ of $G$ defined by $$ \chi(x) = \psi (x N). $$