Lie bracket is a connection? In Road to Reality, section 14.6 on Lie derivative Penrose writes:

Now $\epsilon^2 [j,h]$ corresponds to an
  $O(\epsilon^2)$ gap in the ‘parallelogram’ whose initial sides are $e_j$ and $e_h$ at
  the origin I. The relevant notion of ‘parallelism’ comes from the group
  action, supplying the needed notion of ‘parallel transport’,
  which actually gives a connection with torsion but no curvature.[14.17]

and poses the the exercise:

[14.17] Try to explain why there is torsion but no curvature.

I find it surprising that one could get a connection just from the Lie bracket,
since connections depend on additional structure like the metric tensor,
while Lie brackets do not.
One forum has a proposed answer
giving the Lie bracket itself as the connection $\nabla_{L}M = [L, M]$
if I understood it correctly.
The proof proposed there convincingly shows that there would be torsion & no curvature,
but the proposed answer does not seem to be a connection in the first place
since multiplying the the vector fields $L$ and $M$ by a scalar field $\phi$  does not satisfy a condition required given earlier in the book:


*

*linearity in $L$ : $\nabla_{\phi L}M = \phi\nabla_{L}M$


Treating veactors as directional derivative operators on scalar field shows how
linearity in $L$ fails
$$\begin{align} [\phi L, M](\psi) &= \phi L(M(\psi)) - M(\phi L(\psi)) \\
   &=  \phi L(M(\psi)) - (M(\phi)L(\psi) + \phi M(L(\psi))) \\
   &= \phi[L, M](\psi) -  M(\phi)L(\psi)
\end{align}$$
because of the additional $M(\phi)L(\psi)$ term.
Also if it is possible to define a connection just from the Lie bracket,
why is it usually said that you need additional structure to define parallel transport? Is it just that the all torsion no curvature property stops this connection from being useful, so people disregard it?
So what is Penrose getting at?
Am I missing something?
EDIT
From the answers it seems that the Lie bracket really is a connection, just not a linear or affine one. The book does not formally define what a connection is in general or specify the type of connection is intended when the term is used without qualification. It only gives the derivative style algebraic laws connections a connection must satisfy, including linearity wrt $L$ above. So the Lie bracket does seem to be not a connection in the sense the term used in the rest of the book. That is what I was trying to figure out
I am still curious why it is so frequently claimed that connections, parallel transport and covariant derivatives (which I in my understanding are equivalent concepts) require extra structure on a manifold while Lie derivatives do not, if the Lie bracket is a connection and this connection is hardly ever mentioned.
Also is there a name for the Lie derivative as a connection? Lie connection??
 A: Te connection is indeed $\nabla_LM=[L,M]$ if $L$ is left-invariant; notice that if you have a vector at some point of your Lie group, it can be uniquely extended to a left invariant vector field, so the formula makes sense and defines a connection (btw $\phi L$ is not left-invariant (unless $\phi$ is constant)).
The torsion is (for left-inv. vect. fields) $T(K,L)=[K,L]-[L,K]-[K,L]=[K,L]$, so it's not $0$.
The curvature, on the other hand, is $R(K,L,M)=[K,[L,M]]+\dots=0$ by Jacobi identity. Another way to see that $R=0$ is that any right-invariant vect. field $M$ satisfies $\nabla M=0$, as left and right-inv. fields commute.
A: The parallel transport Penrose is talking about is given by translation in the group. More precisely, the parallel transport $\Gamma(\gamma)^t_s\colon T_{\gamma(s)}G \to T_{\gamma(t)}G$ alongside a curve $\gamma$ is simply the differential of the left translation by $\gamma(t)\cdot \gamma(s)^{-1}$.
We can retrieve the connection from this as follows: Let $\gamma$ be a smooth curve, $X = \dot\gamma(0)$, and $V$ a vector field over $\gamma$ then $$\nabla_X V = \frac{\mathrm d}{\mathrm d t} \Gamma(\gamma)_t^0 V_{\gamma(t)} \bigg\vert_{t=0}.$$
It is now easy to see that $\nabla_XV = 0$ for a left-invariant vector field $V$ and $\nabla_XW = [X,W]$ for a right-invariant vector field $W$. Moreover, as this parallel transport is independent of the curve connecting two points of $G$, the curvature of its associated connection must vanish.
Of course we can define parallel transport by right translation as well and obtain a second connection on $TG$. These two connections agree if $G$ is commutative.
