Can I multiply an element of a lie group with a vector of the tangent space at some point of the Lie group? My lecturer said I CANNOT multiply an element of a Lie group with a vector of the tangent space at some point of the Lie group. Why is that, it seems to me that he does it all the time, for instance
If $G$ is a Lie group, $e$ the identity element of the Lie group,
$Z \in T_eG$ to differentiate $\exp(tZ)$ at $t=0$ usually in my notes they just write without details that the result is $Z$. I think the details are these:
$\frac{d}{dt}|_{t=0}\exp(tZ)=Z\exp(tZ)|_{t=0} = Z \tag{1}$
In the last equality I have a multiplication of an element of the Lie group $\exp(tZ)$ with a tangent vector $Z$, so this contradicts what the lecturer said. So either the lecturer is wrong or this apparently simple derivation is not what I think
Is this correct? Why is the lecturer's statement true(or false)?
 A: The issue is more elementary than you might expect.
The function $L_g : G \to G$ defined by the formula $L_g(h)=gh$ is a diffeomorphism (which follows from the definition of a Lie group). That function therefore has a derivative, namely a smooth map $DL_g : TG \to TG$ defined on the tangent bundle $TG$, which restricts to a linear isomorphism from the tangent space $T_h G$ to the tangent space $T_{gh} G$, for each $h \in G$.
So when one speaks informally about "multiplying an element $g \in G$ by a vector $v$ in the tangent space", although your lecturer is  correct, one can nonetheless interpret the statement as meaning "applying the function $DL_g : TG \to TG$ to $v \in TG$".

Let me add a few words about your own calculations. I don't see any formal validity to your first equation. It does sort of look like a naive extension of a Calculus 1 derivation $\frac{d}{dt} \bigm|_{t=0} e^{kt} = k e^{kt} \bigm|_{t=0} = k$. But if that's what you meant then I'm sure you'll understand that you can't expect a Calculus 1 formula to apply in this context, despite how "naturally" it seems to come out.
Imagine the chaos unleashed on the world if one naively and "naturally" wrote $\exp(W)\exp(Z) = \exp(W+Z)$, in contrast to the actual result for $\exp(W) \exp(Z)$ given by the Baker-Campbell-Hausdorff formula.
A: When you are writing $\exp(tZ)$ you are thinking of the function $f:t\in(-\epsilon,\epsilon)\mapsto\exp(tZ)\in G$. By definition, this is function is the unique solution of the differential equation $u'(t)=Z_{u(t)}$ with initial condition $u(0)=e$ (I am viewing the elements of the Lie algebra as left invariant vector fields), so by definition $$\frac{\mathrm d}{\mathrm dt}\Bigl|_{t=0}f(t)=f'(0)=Z_e.$$ So the equality you want to prove is actually part of the definition of $\exp$.
