Real characters of odd degree

Question

Suppose that $G$ is a finite group and $\chi$ is an irreducible real character, namely that $\chi(g) \in \mathbb{R}$ for every $g \in G$. Is it true that if $\chi(1)$ is an odd number, then $\chi$ is afforded by a representation over real numbers?

Motivation, let $m_{\mathbb{R}}(\chi)$ the Schur index of $\chi$ as defined in $10.1$ of Isaacs' "Character Theory of Finite Groups". If $\mathcal{X}$ is a complex representation that affords $\chi$, then there is an $\mathbb{R}$-representation $\mathcal{Y}$ of twice the dimention of $\mathcal{X}$ such that, viewed as complex, its character is $2\chi$. Hence the Schur $m_{\mathbb{R}}(\chi)$ should be $\le 2$. Nevertheless, by point $h)$ of Corollary $10.2$ of Isaacs' book, we have that $m_{\mathbb{R}}(\chi)$ divides $\chi(1)$ that is odd.

Answer 1

I think I found a quicker, more direct proof. By Brauer-Speiser Theorem (Isaacs' book page 171) we have that $m_{\mathbb{Q}}(\chi)\le 2$ and hence $m_{\mathbb{Q}}(\chi)=1$, since it divides $\chi(1)$ that is an odd number. By point $f)$ of 10.2 of the same book, $m_{\mathbb{R}}(\chi)$ divides $m_\mathbb{Q}(\chi)$.

Answer 2

You don't need Brauer-Speiser or anything complicated, just the Frobenius-Schur indicator and some linear algebra. Since $\chi$ is a real character, the indicator is either $+1$ or $-1$, not $0$. $+1$ means it is a real representation, $-1$ means it is a quaternionic representation. But quaternionic representations preserve an alternating form, and this requires the dimension to be even. So all odd-dimensional (irreducible) representations are of $+$ type, i.e., are real representations.