Show that the trivial representation is the only irreducible complex representation of $G$ defined over $\mathbb{R}$

Question

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.

Answer 1

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.

Answer 2

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 $0$.