Group G hasn't all conditions of Lie group Is there a group $G$ with the property that $G$ is a smooth manifold, the multiplication map of $G$ is smooth, but the inversion map of $G$ is not smooth?
 A: Every Čech-complete paratopological group is a topological group. That means that for Čech-complete groups you do not have to require the continuity of the inverse, continuity of multiplication suffices. Every manifold is Čech-complete. Using the affirmative answer to Hilbert’s fifth problem we get that every paratopological group on a manifold is actually a Lie group uniquely determined by the topological group structure.
In the spirit of Martin let me give a correct (I hope I have not forgotten anything) definition which is even wronger than the definition without the inverse:

A topological space $G$ with a function $\cdot\colon G^2\to G$ is called an $n$-dimensional Lie group if and only if
  
  
*
  
*$G$ is second-countable
  
*There exists an injective, open continuous map $\iota\colon \mathbb{R}^n\to G$
  
*For every $g\in G$ the map $x\mapsto g\cdot x$ is continuous and surjective
  
*There exists $e\in G$ such that $x\mapsto x\cdot e$ is the identity
  
*For every $g\in G$ the map $x\mapsto x\cdot g$ is continuous
  
*$\cdot$ is associative

A: Robert L. Bryant "An Introduction to Lie Groups and Symplectic Geometry" requires in the definition of a Lie group only that the multiplication map be smooth, and then proves that the inversion map must be smooth also.  (Proposition 1, page 14.)
