subgroup and group homomorphism Let $A$ be a group. Then we choose that $B$ is a subgroup of $A$. 
What are the conditions that there exists a group homomorphism from $$A \to B$$
such that the map is surjective (the map is epimorphism)? 
(1) One can consider just finite groups.
(2) How about more general groups? (Lie groups)
 A: What about $\varphi:A\to B$ defined by $\forall x\in A,\varphi(x)=e$ where $e$ is the neutral element of $B\subset A$. One has $\varphi(e)=e$ and $\varphi(x\cdot y)=e=e\cdot e=\varphi(x)\cdot \varphi(y)$. So there must be some more conditions on the homomorphism we're looking for. By the way the above is valid in any group, finite or not.
Now if we want $\varphi$ to be surjective, a simple counterexample is given by $\mathfrak{S}_3$ the group of permutations of $3$ elements.
$$\mathfrak{S}_3=\{e,(1\,2),(1\,3),(2\,3),(1\,2\,3),(1\,3\,2)\}$$
Consider the subgroup $\mathfrak{T}=\{e,(1\,2)\}$, it is impossible to find an epimorphism from $\mathfrak{S}_3$ onto $\mathfrak{T}$ 
A: I believe what you are getting at is the notion of a semidirect product.
That is, if $A$ is a group with subgroup $B$ and normal subgroup $N$ such that:


*

*$B\cap N=1$, and

*$A=BN$,


then $A/N\cong B$ in a very natural way; that is, there is a natural map $A\rightarrow B$. (Note that this is viewing $B$ as a subgroup of $A$, not as an abstract group.)
By "a very natural way", I mean that if $\alpha: B\hookrightarrow A$ is the embedding of $B$ in $A$ and $\phi: A\rightarrow A/N$ is the natural map then the composition $\alpha\circ\phi: B\rightarrow A/N$ is an isomorphism. (This is equivalent to the above properties for groups; we way that the map $\phi$ splits.)
