In characteristic zero, the group algebra is semisimple, so there are finitely many simple representations. These representations correspond to the blocks in the decomposition as a product of matrix algebras. As for characteristic p, this is modular representation theory, The number of irreducibles is the number of $p$-regular conjugacy classes (where $p$-regular means the period is prime to $p$). See, e.g., Serre's Linear Representations of FInite Groups.

The following answer is for finite groups.

In characteristic zero, the group algebra is semisimple, so there are finitely many simple representations. These representations correspond to the blocks in the decomposition as a product of matrix algebras.

Here is a way to determine the number: Start with the field $K$, adjoin the $g$th roots of unity ($g = |G|$) to get $L$, and consider the Galois group $\Gamma_K=L/K$. This is a subgroup of the multiplicative group of the integers mod $g$. Then $\sigma_t \in \Gamma_K$ corresponding to $t \in (\mathbb{Z}/g\mathbb{Z})^*$ acts on $G$ by raising $x \in G$ to the $t$-th power. The dimension of the space of class functions constant on $\Gamma_K$-orbits is the number of simple $K$-representations.

As for characteristic p, this is modular representation theory, The number of irreducibles is the number of $p$-regular conjugacy classes (where $p$-regular means the period is prime to $p$), when the field contains the $g$th roots of unity for $g$ the order of the group. See, e.g., Serre's Linear Representations of FInite Groups. My guess is that it should be true even without the assumption on the field being sufficiently large.