Let $G$ be a finite group of odd order. Prove that if $V$ is a $\mathbb{C} G$-module whose character is real, then dim$V$ and dim$V^G$ have the same parity.
where $V^G:=\{v \in V \mid vg=v \forall g \in G\}$ and having the same parity means they are both odd or both even.
What I have got so far: Since $G$ is of odd order, we can show $g$ is not conjugate to $g^{-1}$ for all $g \in G\setminus \{1\}$. It follows that there is only one irreducible character which is real (i.e. the trivial character).
So if $V$ is an irreducible character, then its character is the trivial character. That is , it is the trivial $\mathbb{C} G$-module, so $V^G=V$ in this case and the result follows.
But if $V$ is reducible, then $1<dim V^G <dim V$. How can I show, in this case that they have the same parity?
First note that, since $\#G$ is odd, we can invert it mod 2: $$\require{cancel} \begin{align*} \dim V\equiv\dim V^G&\iff\chi_V(1)\equiv\langle\chi_V,1_G\rangle=\frac1{\#G}\sum_{g\in G}\chi_V(g)\\ &\iff\cancelto{1}{\color{red}{\#G}}\cdot\chi_V(1)\equiv\sum_{g\in G}\chi_V(g)\\ &\iff\sum_{g\neq 1}\chi_V(g)\equiv 0 \end{align*} $$ So it suffices to show $$s:=\frac12\sum_{g\neq 1}\chi_V(g)$$ is an integer.
Recall $\chi_V(g)$ is an algebraic integer (it is sum of roots of unity), and $\chi_V(g^{-1})=\overline{\chi_V(g)}$. Since $\chi_V$ is real, $\chi_V(g)=\chi_V(g^{-1})$ so summing one representative from each $\{g,g^{-1}\}$, we get $s$ is an algebraic integer. On the other hand, $$s=\frac12[(\#G)\cdot\dim V^G-\dim V]\in\mathbb{Q}.$$ So $s\in\mathbb{Z}$ since $\mathbb{Z}$ is integrally closed.