# Odd-order p-group implies every irreducible representation has Schur index one

Suppose $p$ is an odd prime and $G$ is a finite $p$-group, i.e., a group of prime power order where the underlying prime is $p$. Then, every irreducible representation of $G$ over a splitting field has Schur index (?) 1, i.e., every irreducible representation can be realized over the field generated by the character values of the representation.
• The statement is not true for $p = 2$. An example is the faithful irreducible representation of quaternion group, which is a degree two representation whose character takes values in $\mathbb{Q}$ but where the representation cannot be realized over $\mathbb{Q}$ but is realized over a suitable imaginary quadratic extension of $\mathbb{Q}$, such as any extension of the form $\mathbb{Q}(\sqrt{-m^2 - 1})$.