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?

  • 1
    $\begingroup$ Well, if the character isn't linear, then its value at the identity of the group is $\geq 2$ and couldn't possibly be the identity of $\Bbb C^\times$, so there's that :) $\endgroup$
    – pjs36
    Oct 11, 2015 at 20:15
  • $\begingroup$ Sorry, I do not understand? First why is its value at the identity $\ge 2$, and further why does every linear character arises as an irreducible character? EDIT: Ok, your remark why $\ge 2$ is clear, as at the identity the characters equals the dimension of $V$, but still why every linear character is among the irreducible ones... $\endgroup$
    – StefanH
    Oct 11, 2015 at 20:29
  • $\begingroup$ I get the impression that those notes don't really talk about what it means for a representation to be reducible, merely that there are certain representations that we care about and call irreducible (at least that's what I saw, browsing). If this is the case, it might help to find a more comprehensive source, like Isaacs's Character Theory of Finite Groups. $\endgroup$
    – pjs36
    Oct 11, 2015 at 20:54

1 Answer 1


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


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