At my university, we came across following exercise:
Let $G$ be a finite group of odd order.
Show that every irreducible representation of $G$ over $\mathbb{C}$ with real valued character is trivial (so the trivial representation is the only irreducible complex representation of $G$ defined over $\mathbb{R}$.)
I tried to prove this, but didn´t get very far. As far as I understand, If we take arbitrary irreducible representation, then sum of elements on the main diagonal of the representation matrix must be zero, since character corresponds to the trace of given representation.
I also know irreducible representations correspond to simple modules, but am not sure whether this could be used.
How would you proceed with this? Any idea would be appreciated, thank you.
Best Answer
It’s well known (over $\mathbb{C}$) that there is a perfect pairing $C \times F \rightarrow \mathbb{C}$, where $C$ is the vector space generated by the conjugation classes (denote as $C_G$ their set – a basis of $C$) of $G$, and $F$ is the space of conjugation-invariant functions on $G$, and $F$ has the set $B_F$ of characters of irreducible representations as a basis.
Now, $i_1: \chi \longmapsto \overline{\chi}$ is an involution of $F$, and its ”dual” under the pairing from above is $i_2: [x] \longmapsto [x^{-1}]$ (where $x\in G$ and $[x]=\{gxg^{-1},\, g \in G\}$). So $i_1,i_2$ are symmetries and they have the same number of fixed vectors on $B_F$ and $C_G$ respectively (since it is their trace).
Therefore, one just needs to prove that if $x \in G$ is not the unit, $x$ and $x^{-1}$ aren’t conjugate. But if $x^{-1}=gxg^{-1}$, then $g^2xg^{-2}=x$ so $g^2$ commutes with $x$. As $|G|$ is odd, $g$ is a power of $g^2$ so $g$ commutes with $x$ and thus $x=x^{-1}$, so $x^2$ is the unit. As $|G|$ is odd, it follows that $x$ is the unit. QED.