Show that the trivial representation is the only irreducible complex representation of $G$ defined over $\mathbb{R}$ 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.
 A: 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.
A: Since $\chi$ is real, we have $\langle \chi,\chi\rangle=1$. Notice that
$$\langle \chi,\chi\rangle=\frac{1}{|G|} \sum_{g\in G} \chi(g)\overline{\chi(g)}=\frac{1}{|G|}\left(\chi(1)+2\sum_{g\in I} \chi(g)^2\right),$$
where $I$ is the set of non-trivial elements of $G$ up to inverses. If $\chi(1)$ is even then the sum of the character value squares is half of an integer. But character values are algebraic integers, and this is not possible. Thus $\chi(1)$ is odd. (If you already knew that $\chi(1)\mid |G|$ you don't need this part.)
Now we note that if $\chi(g)\neq 0$ for all $g\in G$ then we are done: the square of a real algebraic integer is at least $1$, so the only way that the sum of $|G|$ many squares of real algebraic integers is equal to $|G|$ is if all have norm $1$, so $\chi(1)=1$.
But the character is the trace of a matrix, so the sum of the eigenvalues. Since $\chi$ is real, the eigenvalues are stable under complex conjugation, and so the non-real eigenvalues come in pairs. Removing those, we are left with an odd number of $1$s and $-1$s, which cannot sum to $0$.
