We let $G$ be a finite group.

If $\chi$ is a complex character of $G$, we define $\overline{\chi}:G \to \mathbb{C}$ by $\overline{\chi}(g)=\overline{\chi(g)}$ for all $g \in G$, and define $\chi^{(2)}:G \to \mathbb{C}$ by $\chi^{(2)}(g) = \chi(g^2)$. We write $\chi_{S}$ and $\chi_{A}$ for the symmetric and alternating part of $\chi$. We note that $\chi_{S}$ and $\chi_{A}$ are characters of $G$ with $\chi^{(2)}=\chi_{S} + \chi_{A}$ and $\chi^{(2)}=\chi_{S} + \chi_{A}$.

First, I want to show that $\overline{\chi}$ is a character of $G$. Now, we can show that $\overline{\chi}(g)=\overline{\chi(g)}=\chi(g^{-1})$ for all $g \in G$ thus $\overline{\chi}(g)$ is a character. Is that OK?

Next, I want to show that $\chi$ is irreducible iff $\overline{\chi}$ is irreducible.

For $(\implies)$ we assume that $\overline{\chi}$ is not irreducible. Thus we must have a reducible representation $\rho:G \to GL(V)$. But then $\chi$ must also be reducible, w.r.t. to that reducible representation, which is a contradiction. $(\impliedby)$ we can show by the same argument. Tbh, it doesn't seem correct to me, I don't think that I really understand what could go wrong here.

Lastly, we let $\chi_{1}$ be the trivial character of $G$. If I understand it correctly, $\chi_{1}(g)=1$ for all $g \in G$. We want to show that $\langle \chi , \overline{\chi} \rangle= \langle \chi_{S},\chi_{1}\rangle + \langle \chi_{A}, \chi_{1} \rangle$.

We have:

$\langle \chi , \overline{\chi} \rangle = \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\overline{\chi(g)} = \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\chi(g^{-1})$

and now I am not sure where to go from here, I'd appreciate any hints.


1 Answer 1


Recall that a character $\chi$ is irreducible if and only if $\langle \chi,\chi\rangle=1$. Note then that

$$\langle \overline{\chi},\overline{\chi}\rangle =\frac{1}{|G|}\sum_{g\in G}\overline{\chi}(g)\overline{\overline{\chi}(g)}=\frac{1}{|G|}\sum_{g\in G}\overline{\chi(g)}\chi(g)$$


$$\langle \chi,\chi\rangle=\frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi(g)}$$

Thus, we see that $\langle \chi,\chi\rangle=\langle\overline{\chi},\overline{\chi}\rangle$.

  • $\begingroup$ Thank you! Somehow, I wasn't aware of this. Is the way of showing that $\overline{\chi}$ is a character correct? $\endgroup$
    – amator2357
    Apr 9, 2019 at 13:10

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