About the elements of a finite subgroup of $\mathrm{GL}(\mathbb{R}^{n})$ Let $G$ be a finite subgroup of $\mathrm{GL}(\mathbb{R}^{n})$. I would like to prove that for every $g \in G$, $\det(g) \in \lbrace -1,1 \rbrace$. 
Here are my ideas : since $G$ is a finite subgroup of $\mathrm{GL}(\mathbb{R}^{n})$, the elements of $G$ satisfy to : $X^{e} - \mathrm{Id} = 0$ (for $e \in \mathbb{N}^{\ast}$). Therefore, the eigenvalues of the elements of $G$ are roots of unity in $\mathbb{C}$. For a given element $g \in G$, we can also note that if $\lambda \in \mathbb{C}$ is an eigenvalue for $g$, then $\overline{\lambda}$ is also an eigenvalue for $g$. Therefore, the determinant of $g$ will be either $-1$ or $1$. Is this correct ?

I'd like to show a somewhat simpler proof, without using eigenvalues theory and stuff.
Since $\det$ is a group homomorphism $\det : \mathrm{GL}(\mathbb{R}^{n}) \to \mathbb{R}^{\times}$ (the codomain being the multiplicative group of reals) the image of $G$ under $\det$ should be a finite subgroup of $\mathbb{R}^{\times}$. In $\mathbb{R}$ only two elements generate finite subgroups: 1 and -1, so it has only two finite subgroups: $\{1\}$ (the trivial one) and $\{1, -1\}$ (isomorphic to cyclic group in two elements). So, the image of $G$ under $\det$ should be one of those two groups, which means that $\det$ has only 1 and -1 as it's values.
