abstract algebra - In what sense are the linear characters among the irreducible characters

Let $G$ be a finite group and denote by $\mathbb C^{\times}$ the multiplicative group of the complex numbers. A linear character is a homomorphism $\chi : G \to \mathbb C^{\times}$. A presentation of $G$ is a homomorphism $\pi : G \to GL(V)$ for some finite-dimensional vector space. If $(\pi, V)$ is a representation of $G$, then the character of $\pi$ is the function $\chi_{\pi}(g) = \mbox{tr} \; \pi(g)$, i.e. the trace of the transformation $\pi(g)$. Now I am reading these lecture notes by Daniel Bump (the linear characters are introduced here), where he says at the end of chapter 2.4 on characters:

f $(\pi, V)$ is a representation, its character $\chi_{\pi}$ is the function $G \longrightarrow \mathbb{C}$ defined by $\chi_{\pi} (g) = \operatorname{tr} \; \pi (g)$. The characters are class functions, meaning that they are constant on conjugacy classes. If $h$ is the number of conjugacy classes of the group, the number of isomorphism classes of irreducible representations of $G$ is also $h$, and if representatives of these are $(\pi_i, V_i)$, let $\chi_i = \chi_{\pi_i}$. These are the irreducible characters of $G$. Among them are the linear characters, which are the homomorphisms $G \longrightarrow \mathbb{C}^{\times}$.

I do not understand the last sentence "Among them are the linear characters, which are the homomorphisms $G \longrightarrow \mathbb{C}^{\times}$." In general for the trace we have $\operatorname{tr}(AB) \ne \operatorname{tr}(A)\operatorname{tr}(B)$, so I guess in general a function $\chi(g) = \operatorname{tr} \; \pi(g)$ could not be a homomorphism $G \to \mathbb C^{\times}$, so how is this statement meant? Or is the trace multiplicative in this special setting?

Answer

The linear characters are precisely the characters (in the trace sense) of $1$-dimensional representations, which are automatically irreducible.