For a matrix Lie group $G\subset GL_n(\mathbb C)$ we define the Lie algebra to be the set of matrices $X\in M_n(\mathbb C)$ such that for all $t\in \mathbb R$ we have $\quad \exp(tX)\in G$.
For clarity, I define the exponential map by $$\exp(X) = \sum_{k=0}^\infty \dfrac{X^k}{k!}$$
The tangent space $T_I(G)$ at the group identity $I$ is the set of all matrices $X\in M_n(\mathbb C)$ for which there exists a differentiable path
$$\gamma : (-\epsilon,\epsilon) \to G$$
where $\gamma(0)=I$ and $\gamma'(0)=X$.
It is proposed that Lie algebra and the tangent space are the same.
It is clear from the differentiability of exponentiation that if $\quad \exp(tX)\in G \quad$ at every $t$ then the path $\gamma$ defined by $\gamma(t) = \exp(tX)$ shows that $X$ is in the tangent space.
The other direction is not as obvious to me. If $X$ is in the tangent space (ie. there exists a path $\gamma$ satisfying the conditions above) then $X$ is in the Lie algebra. Can somebody please explain why this is so?
The following comes from Brian C. Hall's text (2003 edition):$\newcommand{\fg}{\mathfrak{g}}$
The proof (of your direction) is as follows:
Let $\gamma$ be a curve satisfying these conditions. Then, we may state that $\log(\gamma(t)) \in \fg$ for all sufficiently small $t$ (see theorem 2.27: there exists a neighborhood $U$ of $0$ in $\fg$ and a neighborhood $V$ of $I$ in $G$ such that $\exp:U \to V$ is a homeomorphism). Now, define $$ X_{\gamma} = \frac{d}{dt} \log(\gamma(t))|_{t=0} = \lim_{h \to 0} \frac{\log(\gamma(h))}{h} $$ since $X_{\gamma}$ is the limit of elements in $\fg$ and since $\fg$ is closed, $X_{\gamma} \in \fg$. However, we note that (again, for sufficiently small $t$) $$ \log(\gamma(t)) = (\gamma(t) - I) - \frac 12(\gamma(t)-I)^2 + \frac 13(\gamma(t) - I)^3 - \cdots $$ Differentiate term by term by applying the product rule. All terms besides the first approach zero as $t \to 0$. Thus, we have $$ \frac{d}{dt} \gamma(t)|_{t=0} = \frac{d}{dt} \log(\gamma(t))|_{t=0} = X_{\gamma} \in \fg $$ which was the desired result.