We know that if $G$ be a finite group and $F$ be an algebraically closed field whose characteristic does not divide the order of $G$, then the number of inequivalent irreducible $F$-representations of $G$ equals the class number of $G$.
Now, if we suppose that the field $F$ is not necessarily an algebraically closed field but its characteristic does not divide the order of $G$, is it true that the number of inequivalent irreducible $F$-representations of $G$ cannot exceed the class number of $G$?
If the answer is yes, how can we prove that?
I know that each irreducible representation of $G$ over $F$ is completely reducible by Maschke's Theorem because the characteristic of $F$ does not divide the order of $G$ and $G$ is finite. Also, each representation of $G$ over $F$ can be considered as a representation of $G$ over $\overline{F}$, where $\overline{F}$ is an algebraic closure of $F$. But I have no idea about how they can help.
I will be so grateful for any answers and comments.