5
$\begingroup$

I've been given the following two questions, and for both I'm really unsure what to do (I'm a beginner with the theory of characters).

Let $G$ be a finite group, $\mathbb{C}G$ its group algebra over $\mathbb{C}$. $\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$ forms a $\mathbb{C}G$-module with the action defined, for $a,b \in \mathbb{C}G$, $f \in \text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$, by $(af)(b) = f(ba)$.

With this module structure defined on $\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})$, firstly I want to find $\chi_{\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})}(g)$ for all $g \in G$. Secondly, in the case that $G = S_3$, I want to express $\chi_{\text{Hom}_{\mathbb{C}}(\mathbb{C}G,\mathbb{C})}(g)$ in terms of simple characters. Could someone please clearly describe how to do this.

  • 0
    Interesting question, I myself am also currently learning about characters. Don't think I can start a bounty just yet but I would also appreciate it if someone could provide an answer for this.2012-10-17
  • 0
    This is not a clear solution, but something good to keep in mind: Group algebra is selfinjective (in fact symmetric), so $K$-linear dual (here $K$ is an algebraically closed field) of left $KG$-module $KG$ is isomorphic itself as right $KG$-module. Over charactersitic zero, $KG$ is completely reducible (semisimple), hence it is easy to see the characters of the $K$-linear dual is the same as the original one. Over other characteristic, the final result is the same by tweaking the argument using the property that $KG$ is symmetric.2012-10-18

2 Answers 2