Question

Let G = S3, and k be any field of characteristic 3. Show that there are only two irreducible representations for G over k, namely, the trivial representation and the sign representation.

Answer 1

Let $U$ be an irreducible $kG$-module, and let $T$ be the action of (123) on $U$. Since $(T - I)^3 = T^3 - I = 0$, $T$ has an eigenvalue 1, so $U_0 = \{u \in U : Tu = u\}$ is nonzero. For $u \in U_0$, we have (123) $((12)u) = (12)(132)u = (12)(123)^2 u = (12)u$, so (12) $u \in U_0$ also. This shows that $U_0$ is a $kG$-submodule of $U$, so $U_0 = U$, and we may view $U$ as a $kG$-module where $G = G / <(123)> = <(12)'>$. Since (12) has order 2, we have $U = U_+ \oplus U_-$ where $U_+ = \{u \in U : (12)u = u\}$, and $U_- = \{u \in U : (12)u = -u\}$. Therefore, we have either $U = U_+$ (giving the trivial representation), or $U = U_-$ (giving the sign representation).