Proof: Matrix exponential maps from tangent space to Matrix (Lie) group

Question

Let us assume we have a definition of the tangent space (e.g. as in Proof: Tangent space of the general linear group is the set of all squared matrices). Furthermore, we already verified that the tangent space $\mathfrak{gl}(n)$ of the general linear group $GL(n)$ -- the group of invertible matrices -- is the set of all square matrices $\mathbb{R}^{n\times n}$.

Afterwards, we introduce the matrix exponential: \begin{equation} \exp(\mathtt{X}):\mathbb{R}^{n\times n}\rightarrow \mathbb{R}^{n\times n},\quad \exp(\mathtt{X})= \sum_{k=0}^\infty\frac{\\mathtt{X}^n}{n!}~. \end{equation} We can prove that $\exp(\mathtt{X})$ is invertible [indeed $\exp(\mathtt{X})^{-1}=\exp(\mathtt{-X})$], thus $\exp(\cdot)$ maps element from the tangent space $\mathfrak{gl}(n)$ to the general linear group $GL(n)$.

Now, we are interested in all subgroups of $GL(n)$, and we will simply call them matrix Lie groups $G\subset GL(n)$:

How can we show in general that $\exp(\cdot)$ maps elements from the tangent space $\mathfrak{g}$ to the corresponding matrix Lie group $G$?

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Formal:

Let G be a subgroup of $GL(n)$ and $\mathfrak{g}$ be the tangent space of $G$.

Show: $x\in \mathfrak{g} \Rightarrow \exp(x)\in G.$

Answer 1

Let $\gamma$ be the integral curve in $G$ of the left invariant vector field associated to $X$, i.e. $V(Y) = YX$. This means $\gamma(0) = e$ and $\gamma'(t) = \gamma(t) X$ for $Y \in G$. Then $\gamma$ is unique (even as considered a curve in $GL(n)$) by theorems of ODE but it is easy to see that the curve $\exp(tX)$ satisfies the defining properties of $\gamma$ so must be $\gamma$. In particular, $\exp X \in G$.