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.