Finite Rotation Groups One can identify if a given finite group is a subgroup of $ O_n(\mathbb{R}) $ by checking its character table for faithful $ n $ dimensional + type irreps.
Is there any way to figure out which finite subgroups of $ O_n $ are rotation groups (i.e. subgroups of $ SO_n $) just based on character tables or something similar?
Certainly any subgroup of $ O_n $ which is not a subgroup of $ SO_n $ must have an index 2 normal subgroup.
 A: Here's how to show that the character $\chi_V$ of a representation $V$ determines its determinant. This is essentially an exercise in symmetric functions. To say it in as low-tech a way as possible, for $g \in G$, if you know all character values $\chi_V(g^k)$ then you know all power sums $\sum \lambda_i^k$ of the eigenvalues of the action $\rho_V(g)$ of $g$ on $V$, which means you know all these eigenvalues (this is the exercise in symmetric functions bit), which means you know the product $\det(\rho_V(g))$ of these eigenvalues. A slick way to package the symmetric function identity we need here is
$$\det(I + \rho_V(g) t) = \exp \left( \sum_{k \ge 1} (-1)^{k-1} \frac{\chi_V(g^k) t^k}{k} \right)$$
and then you can recover $\det(\rho_V(g))$ itself by computing the coefficient of $t^{\dim V}$ on both sides. This identity follows from writing $\det(I + \rho_V(g) t) = \prod_{i=1}^{\dim V} (1 + \lambda_i t)$ where $\lambda_i$ are the eigenvalues, then taking the logarithmic derivative of this product.
The coefficient of $t^k$ on the LHS is the character of the $k^{th}$ exterior power $\Lambda^k(V)$ so this argument shows more generally that the character of a representation determines the characters of its exterior powers. A similar identity gives the characters of the symmetric powers.
