Real representation of group of odd order

Question

Let $G$ be a finite group of odd order. Suppose that $G$ has a real 4-dimensional faithful representation. Is it true that $G$ should be abelian in this case?

Answer 1

We can assume $G\leq O(4)$ wlog. Because $|G|$ is odd, we have in fact $G\leq SO(4)$ ($g\mapsto g^2$ is a bijection and hence every $g$ must have positive determinant) and a monomorphism $G\to PSO(4) \cong SO(3)\times SO(3)$. Now lets look at the two projections $\pi_1(G), \pi_2(G)\leq SO(3)$. We consult the list of finite order subgroups of $SO(3)$ and find that the only odd-order subgroups of $SO(3)$ are the cyclic ones. Therefore $G\leq \pi_1(G)\times\pi_2(G)$ is abelian.

Answer 2

Alternatively, a finite group of odd order has no real-valued complex irreducible character other than the trivial character. ( Proof (well-known): if $\chi$ is such a character, we have $\chi(g) = \chi(g^{-1})$ for all $g \in G$ and $g \neq g^{-1}$ whenever $g \neq 1_{G}.$ Hence $ \chi(1) + 2 \alpha = 0$ for some algebraic integer $\alpha$ by the orthogonality relations, a contradiction as $\chi(1)$ is odd). So if we regard the $4$-dimensional real representation as a complex representation, all non-trivial irreducible constituents of its character must occur in complex conjugate pairs. Since $G$ has no complex irreducible character of degree $2,$ we see that all irreducible constituents of the given character must be linear (ie degree $1$). Hence $G$ is Abelian, as the given character is assumed faithful. Later remark: the statement remains true for $5$-dimensional faithful representations of finite groups of odd order, but there is a non-Abelian group of order $27$ with a faithful $6$-dimensional real representation.