Definition of Lie Groups In the definition of Lie Group, we require that $$(x,y)\rightarrow x*y \text{ and } x\rightarrow x^{-1}$$ both be smooth. Are there any examples of groups that satisfy only one of these and not the other and are hence are not smooth manifolds? The definition I am using of smooth manifold is the same as we use for topological spaces, i.e. if it is locally diffeomorphism to $\mathbb{R}^n$
The reason I am asking this is because, I am wondering if the latter requirement follows from the former.
Edit:
Are there any examples of groups that satisfy only one of these and not the other and are hence are not smooth manifolds?
should be:
Are there any examples of groups that satisfy only one of these and not the other and are hence are not Lie groups?
 A: Here's a set-theoretic family of examples of smooth manifolds that are groups with smooth inversion but not multiplication. Let $(G,e,*)$ be a group of exponent $2$, i.e. $g *g=e$ for every $g\in G$ with the cardinality $\mathfrak{c}$ of $\mathbb{R}$. For instance, $G$ could be a direct product of $\mathfrak{c}$ $\mathbb{Z}_2$s. Let $\phi:\mathbb{R}\to G$ be any bijection and define a group structure on $\mathbb{R}$ by $x\star y=\phi^{-1}(\phi(x)*\phi(y))$. This kind of construction always yields a group. Now since we picked our $G$ to have an inversion map preserved under bijection, inversion is guaranteed to be smooth: for $x\in\mathbb{R}, x^{-1}_\star=\phi^{-1}(\phi(x)^{-1}_*)=\phi^{-1}\phi(x)=x$, i.e. the inversion map $\mathbb{R}\to\mathbb{R}$ is just the identity.
But for the vast majority of choices of $\phi$ the multiplication will not be smooth. For since $\mathbb{R}$ has $\mathfrak{c}^\mathfrak{c}$ self-bijections, we've exhibited $\mathfrak{c}^\mathfrak{c}$ distinct group structures on $\mathbb{R}$ all with smooth inversion. On the other hand, a continuous-in particular smooth-group structure on $\mathbb{R}$ is specified by the maps $x,y\mapsto x\star y$ for $x,y\in \mathbb{Q}$, i.e. by an element of $\mathbb{R}^{\mathbb{Q}\times\mathbb{Q}},$ which has cardinality only $\mathfrak{c}$!
Edit Incidentally, smoothness of multiplication actually implies smoothness of inversion if inversion is continuous. I don't know if there are groups with smooth multiplication and discontinuous inversion-my guess is there are. Re-edit But as Jack Lee's comment below shows, in fact smooth multiplication does imply smooth inversion.
