Prove that the one-dimensional characters of a group $G$ form a group under multiplication of functions. This group is called the character of $G$. Here's my try.
Let $g$ be an element of order $k$ present in some conjugacy class. Let $X(g)$ denote the character of an element in the same conjugacy class as $g$. Since all the characters we're concerned with are one-dimensional, all their values on one conjugacy class are powers of the k-th root of unity. Now a group has to be closed under multiplication and taking inverses: $$X_1\left(g\right)X_2\left(g\right)=e^{\frac{2\pi i}{k}t_1}e^{\frac{2\pi i}{k}t_2}=e^{\frac{2\pi i}{k}\left(t_1+t_2\right)}$$ and so in order for the group to be closed under multiplication the element $e^{\frac{2\pi i}{k}\left(t_1+t_2\right)}$ has to be the value some character $X_3$ takes on the conjugacy class represented by the element $g$. Similarly taking inverses translates to the existence of a character which attains the value $e^{\frac{-2\pi i}{k}t}$ on the same conjugacy class.
But how can I be sure that for each $t_1$ and $t_2$ (powers of the k-th root of unity) there exists a one dimensional character which attains the values of the powers $t_1+t_2$ (for closure under multiplication) and another that attains the value $-t$ (for closure under taking inverses).
Best Answer
$u: G \to \Bbb{C}^*$ is an homomorphism iff $u(gh) = u(g)u(h)$. Let $w(g) = u(g)v(g)$ then $w(gh) = u(gh)v(gh)= u(g)u(h)v(g)v(h) = w(g)w(h)$.
Same for $z(g) =1/u(g)$.
If $G$ is a finite group then $u(g)$ is a $|G|$-th root of unity.
The character table is something different : it is the traces of the finite dimensional (irreducible) representations, and they don't form a group (nor do the representations). Since $tr(AB)= tr(BA)$ we obtain $tr(\rho(ghg^{-1})) =tr(\rho(g^{-1})\rho(gh))= tr(\rho(h))$.
For the homomorphisms we have much more : since $\Bbb{C}^*$ is abelian we obtain $u(ghl) = u(hgl)$ thus $u(g) = u(h)$ whenever $g h^{-1} \in [G,G]$ (the commutator subgroup).
$G^{ab} = G/[G,G]$ is abelian and the characters of $G$ is the same as the characters of $G^{ab}$. If $G$ is finite or finitely generated then so is $G^{ab}$ and it is easy to find all its characters.