This is a question from Lee's book:
The definition of the exponential map in terms of the one parameter subgroups seemed quite abstract, so I think that fully understanding this exercise will be helpful. As far as I understand, if we find the one parameter subgroups (that is to say: the maximal integral curves of left-invariant vector fields which start at $e$), we're done. So we want to solve $$ \cases{\gamma'(t)=X\\\gamma(0)=e} $$ For $X\in \textrm{Lie}(G)$. In the case $G=\mathbb{R}^n$, $\textrm{Lie}(G)\cong \mathbb{R}^n$ so I guess we get $\gamma(t)=Xt$. What about the torus? Maybe I can use that $T^n\cong S^1\times\ldots\times S^1$, but how?
Update:
What I did is not really correct: I should have written
$$ \cases{\gamma'(t)=X_{\gamma(t)}\\\gamma(0)=e} $$ Now, $X_{\gamma(t)}=L_{\gamma(t),*}(e)X_e$. But now I do not know how to justify that this is simply $L_e$. i.e. how do I determine $L_{\gamma(t),*}(e)$?
I tried the following: $L_{\gamma(t)}(x)=\gamma(t)+x$ (with $x\in\mathbb{R}^n$). Hence the matrix for $L_{\gamma(t),*}(e)$ is $$ \left(\begin{array}{ccc} \frac{\partial L_{\gamma(t)}^{1}}{\partial x^{1}} & \ldots & \frac{\partial L_{\gamma(t)}^{1}}{\partial x^{n}}\\ \vdots & & \vdots\\ \frac{\partial L_{\gamma(t)}^{n}}{\partial x^{1}} & \ldots & \frac{\partial L_{\gamma(t)}^{n}}{\partial x^{n}} \end{array}\right) = I $$ and therefore $L_{\gamma(t),*}(e)X_e=X_e$. So now only the torus is left.
See $S^1$ as $\{z\in\mathbb{C}\,|\,|z|=1\}$. If $X=(t_1,\ldots,t_n)$, consider the map $\gamma\colon\mathbb{R}\longrightarrow\left(S^1\right)^n$ defined by $\gamma(x)=\bigl(e^{ixt_1},e^{ixt_2},\ldots,e^{ixt_n}\bigr)$. It is clear that $\gamma$ is a homomorphism and that $\gamma'(0)=X$. Therefore,$$\exp(X)=\gamma(1)=\bigl(e^{it_1},e^{it_2},\ldots,e^{it_n}\bigr).$$