I'm trying to show that if $G$ is a finite odd order group then, all of its nontrivial complex representations are of complex type (i.e., it is not realisable over the reals).

(I have answered it here: If $G$ is a finite non-trivial group of odd order, it has an irreducible representation not realisable over the reals.)

Let $V$ be such a representation, and let $V^G:=\{v\in V: gv=v, \forall g \in G\}$ be its fix point set. Using some theorems in Bröcker's book about Representations of Compact Lie Groups, I can solve this problem if I show that $V^G=0$. Is that true?

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    $\begingroup$ Well, what about the trivial representation? $\endgroup$ – Matt Samuel Jan 14 '19 at 0:11
  • $\begingroup$ It should be a non-trivial representation*, I'll change @MattSamuel $\endgroup$ – Andre Gomes Jan 14 '19 at 0:12
  • $\begingroup$ I have answered the first question here: math.stackexchange.com/questions/1033844/… $\endgroup$ – Andre Gomes Jan 14 '19 at 17:24

Any nonzero vector $v\in V^G$ spans a $1$-dimensional trivial sub-representation. If $V$ is irreducible of dimension $>1$, then it follows that $V^G=0$. If $\dim V=1$, then $V^G\neq 0$ implies $V$ is the trivial representation.

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