I am trying to prove the following problem (Isaacs "Character Theory of finite groups" 2.6):
Let $\chi, \psi$ be characters of $G$. Define $\chi\psi : G \rightarrow \mathbb{C}$ by $(\chi \psi)(g) = \chi(g)\psi(g)$.
a) If $\psi(1) = 1,$ show that $\chi\psi$ is a character.
b) If $\psi(1) = 1,$ show that $\chi\psi \in Irr(G)$ iff $\chi \in Irr(G).$
a) Let $\varphi$ be a representation of the character $\chi$. Let's define the following representation $\stackrel{\sim}{\varphi}:$
$$\stackrel{\sim}{\varphi}: G \rightarrow GL_{n}(\mathbb{C})$$
$$g \stackrel{\stackrel{\sim}{\varphi}}{\rightarrow} \psi(g)\varphi(g)$$
This is really a representation, since $$gh \stackrel{\stackrel{\sim}{\varphi}}{\rightarrow} \psi(gh)\varphi(gh) = \psi(g)\psi(h)\varphi(g)\varphi(h)$$ (since $\psi$ is a linear character and $\varphi$ is a representation).
Then this representation will correspond to the character $\chi\psi$.
Are the arguments given here correct?
b) Suppose that the character $\chi \psi$ is irreducible. Then
$$[\chi \psi, \chi \psi] = \frac{1}{|G|}\sum\limits_{g\in G}|\chi \psi(g)|^{2} = \frac{1}{|G|}\sum\limits_{g\in G}|\chi(g)|^{2} |\psi(g)|^{2} = 1 \geq \frac{1}{|G|}\sum\limits_{g\in G}|\chi(g)|^{2} > 0 \Rightarrow \frac{1}{|G|}\sum\limits_{g\in G}|\chi(g)|^{2} = 1.$$
But how to prove the opposite statement?
I will be very grateful for any help.
Best Answer
Your part (a) is fine. Keep in mind $\psi(g)$ is scalar-valued, which is why it commutes with everything.
For part (b), don't bother with inequalities. Since $G$ is finite and $\psi(g)$ is scalar-valued, if $m$ is the order of $g$ we can say $\psi(g)^m=\psi(g^m)=\psi(e)=1$ so $\psi(g)$ is an $m$th root of unity and in particular, $|\psi(g)|=1$ for all $g\in G$. Thus, when you write out the formula for $\langle\chi\psi,\chi\psi\rangle$, you will find it equals the formula for $\langle\chi,\chi\rangle$, so one is $=1$ iff the other is, so $\chi\psi$ is irreducible iff $\chi$ is.