I have the following exercises:
(a) Verify directly that the group $SO(n)$ has a natural Lie group structure.
(b) Given this structure, prove that $SO(n) \subset GL(n;\mathbb{R})$ is a Lie subgroup.
(c) $GL(n;\mathbb{R})$ lies as an open subset in $M(n;\mathbb{R})$, the vector space of $n \times n$ matrices with real entries. Consequently, the tangent space $T_{e}GL(n;\mathbb{R})$ can be naturally identified with $T_{e}M(n;\mathbb{R}) \cong M(n;\mathbb{R})$. Using this identification, show that $T_{e}SO(n)$ is a subspace of $M(n;\mathbb{R})$.
So far this is what I've shown: First we consider the orthogonal group, $O(n)=(A \in GL(n;\mathbb{R}) : A^{T}A=I)$ where $(A^{T})_{ij}=A_{ji}$. Since we know that $(AB)^{T}=B^{T}A^{T}$, the set of orthogonal matrices forms a subgroup. We also see that $SO(n) \subset O(n)$ as the subgroup of matrices with determinant 1. It remains to be shown that $O(n), SO(n)$ have a manifold structure. We introduce $\mathrm{Sym}(n;\mathbb{R})$ to denote the vector spave of $n \times n$ symmetric matrices. This vector space has dimension $n(n+1)/2$ and is a linear subspace of $M(n;\mathbb{R})$. Thus, we may consider it as a local piece of $\mathbb{R}^{n(n+1)/2}$. We write $\phi: M(n;\mathbb{R}) \rightarrow \mathrm{Sym}(n;\mathbb{R})$ such that $A \rightarrow A^{T}A$. This map is quadratic in the entries of the matrix, so it is a smooth map between Euclidean spaces. Finally, we note that the identity matrix $I \in \mathrm{Sym}(n;\mathbb{R})$ is a regular value of $\phi$. Here we invoke the Implicit Function Theorem to see that $\phi^{-1}(I)$ is a submanifold of $GL(n;\mathbb{R})$. Thus, both $O(n), SO(n)$ have group and manifold structures. Hence, $SO(n)$ has a natural Lie group structure.
I guess I'm a little unsure of the distinction between (a) and (b). What must I further show? Have I given the natural structure or was this an overly complicated method? Finally, any help with (c) would be greatly appreciated.
For (c) let $X(t)$ be a curve in $O(n)$ such that $X(0)=e$. Then we have $X(t)X^T(t)=I$ differentiation gives $X'(0)+X'^T(0)=0$. And vice versa if you have a skew-symmetric matrix $A$ then $e^{tA}$ is a curve with tangent vector $A$.