Can someone point out the gap in this argument. Consider a simplyconnected Lie group with the ()connection. This connection is flat and so the sectional curvatures are zero. Then, by the CartanHadamard theorem and simpleconnectedness, the Lie group must be diffeomorphic to ${\Bbb R}^n$. However, I don't think that this is correct without an addition assumption of solvability or nilpotency. What's wrong here?

As Emerton pointed out, you need to be careful about the connection. CartanHadamard theorem is a statement involving the curvature of the LeviCivita connection determined by some metric. If $G$ is a Lie group equipped with a biinvariant metric $h$, then this metric induces a metric $\langle,\rangle$ on the Lie algebra $T_1G$. Given two unit orthogonal vectors $X,Y\in T_1G$ that span a plane $\pi\subset T_1G$, then the sectional curvature of $h$ at $1$ along the plane $\pi$ is given by $$ K_1(\pi)=\frac{1}{4}\langle\;[x,y],[x,y]\;\rangle$$. We see from the above equality that the sectional curvature everywhere $\leq 0$ iff the Lie algebra is Abelian. For metrics on $G$ which are only left invariant the computations of curvatures are a bit more complicated and you can find more details in John Milnor's paper, Curvatures of left invariant metrics on Lie groups, Adv. in Math., vol. 121(1976), p. 293329. 


Your connection is not Riemannian; it has torsion, so cannot be the LeviCivita connection of any Riemannian metric. The CartanHadamard theorem isn't even true in Lorentzian geometry, and so you wouldn't expect it for a flat connection which isn't torsion free. 

