is the subgroup generated by one-parameter unipotent subgroups a Lie subgroup? Let $G$ be a Lie group and $H$ be a subgroup generated by some
one parameter unipotent subgroups (in group sense). Is it true that
$H$ has a Lie group structure which makes it a Lie subgroup of $G$?
Is $H$ closed? 
How about for general subgroups?
 A: The many comments as well as the answer by zroslav probably add to the confusion resulting from the original unfocused formulation of the question.   First, it's not really a question about Lie groups but about algebraic groups:   In a Lie group there is generally no intrinsic Jordan decomposition, hence no intrinsic notion of unipotent subgroup as seen already in dimension 1.  
The algebraic group theory was mostly shaped by Borel and Chevalley, both of whom were motivated in part by an interest in Lie groups.    But a Lie group is initially a real manifold, whereas algebraic groups start out over algebraically closed fields and then get more complicated relative to smaller fields of definition.   For an algebraic group, taken in the Zariski topology, "irreducible" = "connected".   Here a finite matrix group can be viewed as an algebraic group but is connected only when trivial.    In any case, an affine algebraic group can always be embedded in some general linear group, so over $\mathbb{C}$ you get a Lie subgroup of the complex general linear group (viewed as real).   The point is that being Zariski-closed implies being closed in the usual sense. 
An early result of Chevalley (treated in several books with the title Linear Algebraic Groups, including section 7.5 of my book) shows the importance of conditions on irreducibility for study of a group generated by closed subsets of an algebraic group.   One consequence of Chevalley's theorem is that two closed connected subgroups will generate a closed connected subgroup in a nice way, but the theorem has other applications as well.    All of this comes very early in the theory, before Jordan decomposition and the detailed structure of reductive groups, where you can ask more interesting specific questions about what various subgroups generate.
A: The answer is "yes" for solvable groups: there is a decomposition of any solvable group on its maximal tori and unipotent radical.
For semi-simple groups the answer is "yes": I've read this statement in Humphrey, "Introduction to Lie Algebras and Representation Theory" (H coincides with G).
In the general case the answer is also "yes": there is a Levy decomposition of any Lie group on its maximal reductive subgroup and its unipotent radical.
A: A good reference to Rongguang's questions is Proposition 2.2.6 in Springer's book: Linear Algebraic Groups 2nd Edition.
