# Complex character of a finite group $G$

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.

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 $$\langle \chi,\chi\rangle=\langle\overline{\chi},\overline{\chi}\rangle$$.
• Thank you! Somehow, I wasn't aware of this. Is the way of showing that $\overline{\chi}$ is a character correct? – amator2357 Apr 9 '19 at 13:10