I was reading "Linear Representations of Finite Groups" from Jean-Pierre Serre and in chapter 2.2 he says has the following:
If ϕ and ψ are functions on G, <ϕ,ψ> = $\frac{1}{g}$$\sum_{t∈G} $$\mathrm{ϕ(t}^{-1})$ψ(t)$ =$$\frac{1}{g}$ $\sum_{t∈G} ϕ(t)$$\mathrm{ψ(t}^{-1})$.
We have <ϕ,ψ> = <ψ,ϕ>.
I don't know why <ϕ,ψ> = <ψ,ϕ> is true.
I know that ${\displaystyle \langle ϕ,ψ\rangle ={\overline {\langle ψ,ϕ\rangle }}}$ But i don't know why ${\displaystyle \langle ψ,ϕ\rangle ={\overline {\langle ψ,ϕ\rangle }}}$.
I found out here https://en.wikipedia.org/wiki/Complex_conjugate that ${\displaystyle \varphi ^{2}=\operatorname {id} _{V}\,}$ therefore i have $\mathrm{ϕ^{-1}}$ = ${\overlineϕ}$
And then the result follows from that, but i don't know if this is right.
I'm sorry if my question is off-topic, but i don't know how to address it.
The set $\{t\mid t\in G\}$ is equal to the set $\{t^{-1}\mid t\in G\}$ and therefore\begin{align}\langle\varphi,\psi\rangle&=\frac1{\#G}\sum_{t\in G}\varphi(t^{-1})\psi(t)\\&=\frac1{\#G}\sum_{t\in G}\varphi(t)\psi(t^{-1})\\&=\frac1{\#G}\sum_{t\in G}\psi(t^{-1})\varphi(t)\\&=\langle\psi,\varphi\rangle.\end{align}