Lie Algebra of a lie subgroup of a lie group

Question

So this is a doubt that im having about lie algebras of lie subgroups of lie groups,

Most of the times when we want to prove that something is a lie Group, the hard part its proving its a manifold, for example for matrices, so we create a function from a manifold that cointains our desired set and prove that the function has a regular value and its pre-image is our set.Then we know how to calculate tangent spaces, and then we know how to calculate lie algebras. My question is for example, we have the orthogonal group wich is going to be a submanifold of $M_{n\times n}$, and we know how to calculate its lie algebra. Then imagine i want to see what is the lie algebra of $SO(n)$, so first i gotta prove its fact a lie group, and i create a function $f : O(n) \rightarrow \mathbb{R}$ such that $f(A)=det(A)$ and then we can prove that $1$ is a regular value of this funciton and so $SO(n)$ is going to be a lie group. My question is when calculating the lie algebra of this subgroup i need to calculate $ker(df)_I$ and then intersect it with the lie algebra of $O(n)$ right?

Answer 1

Yes, that is correct. Note that, in general, it is quite easy prove that a given subgroup of a Lie group is a Lie subgroup. By Cartan's theorem, it is a Lie subgroup if and only if it is a closed subset.

Answer 2

In this case, you're working way too hard. $O(n)$ has two connected components — $SO(n)$ is the component containing the identity (all the matrices $A$ with $\det A = +1$, and the rest have $\det = -1$). This means that, in fact, $\mathfrak{so}(n) = \mathfrak o(n)$.