I have a rather basic mis-understanding about Lie groups and Lie algebras. Consider the Lie group $SO(N)$ for $N>3$ of rotations on $\mathbb{R}^N$. On the one hand this Lie group has dimension $N(N-1)/2$, since every $SO(N)$ element can be parametrized as $e^{X}$, where $X$ is an anti-symmetric matrix with $N(N-1)/2$ free parameters.
On the other hand, $\mathbb{R}^N$ has $N$ cardinal axes. Can I not express every rotation in $\mathbb{R}^N$ as products of rotations about the axes, implying that $SO(N)$ has dimension $N$? Where do the additional degrees of freedom come from? A follow up related question is: each Lie algebra generator $X(i,j)$, for $1 \leq i < j \leq N$ generates a one-parameter group of rotations $O(\theta) = e^{\theta X_{ij}}$. Is there a simple geometric explanation for which axis this rotation is about?
Firstly note that already in just 2 dimensions this doesn't make sense. $SO(2)$ represents rotations in the plane and is 1 dimensional. So the naive assumption should be that we have $\frac{N}{2}$ ($\frac{N-1}{2}$ if $N$ is odd) dimensions for $SO(N)$ by splitting into orthogonal 2-dimensional subspaces . The problem being that products of these simple rotations don't generate all possible elements of $SO(N)$. Indeed they generate exactly a Cartan subgroup.
The next naive guess is that we shouldn't have made the 2-dimensional subspaces orthogonal and worked with all subspaces or at least all subspaces we get from a fixed orthonormal basis. But the number of those is exactly $N$ choose $2$ or $\frac{N(N-1)}{2}$ which is the correct answer. Obviously I haven't shown that there is no redundancy there but this is the right idea.