Does the set of diffeomorphisms which are induced by flows form a group? Let $M$ be a smooth manifold. 
Consider the set of diffeomorphisms which are induced by flows of vector fields. (which are not time-dependent)
Is this set a subgroup of $\text{Diff}(M)$?
(Note that not every diffeomorphism which is isotopic to the identity is induced by a flow of a vector field, see here for details).
"A naive attempt":
Maybe it's possible to construct a counter-example when taking $M=\mathbb{S}^2$. Every vector field on $\mathbb{S}^2$ vanishes at some point, hence every flow-diffeomorphism has a fixed point. Maybe we can find two vector fields, such that the composition of their flows is a diffeomorphism without fixed points.
 A: The relevant theorem is 
Theorem (W.Thurston). Let $M$ be a smooth compact manifold, $Diff_o(M)$ is the identity component of the diffeomorphism group of $M$. Then the group $Diff_o(M)$ is simple (as an abstract group). 
You can find a proof in 
A. Banyaga, The structure of classical diffeomorphism groups. Mathematics and its Applications, 400. Kluwer Academic Publishers Group, Dordrecht, 1997.
Given this, let $G< Diff_o(M)$ denote the subgroup generated by the set $F_M$ of diffeomorphisms given by flows of time-independent vector fields on $M$. It is clear that this subgroup is normal and nontrivial (provided that $dim(M)>0$). Hence, $G= Diff_o(M)$ by Thurston's theorem. It follows that the set $F_M$ cannot form a subgroup unless $F_M= Diff_o(M)$. But, as you already know (see also Lee Mosher's answer here in the case of surfaces), $F_M\ne Diff_o(M)$. Therefore, $F_M$ does not form a subgroup. 
Edit. For a direct proof that $F_M$ is not a subgroup see the answer by Martin M.W. to this MO question. 
