Question

Let k be a field whose characteristic is prime to the order of a finite group G. Show that the following two statements are equivalent:
(a) each irreducible kG-module has k-dimension 1;
(b) G is abelian, and k is a splitting field for G.

Answer 1

Assume (a). Then we have $1 = \dim_k M_i = n_i \dim_k D_i$, which implies that $n_i = 1$ and $\dim_k D_i = 1$. Therefore, $D_i = k$, and we have $kG \preceq k \times \dots \times k$ (since $\operatorname{rad} kG = 0$ here). Clearly, this gives (b). Conversely, if (b) holds, then $n_i = \dim_k D_i = 1$ for all $i$, and we have $\dim_k M_i = n_i \dim_k D_i = 1$, as desired.