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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)

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    Always. (Did you forget an extra condition?) – 2017-02-17
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    I am not sure I understand your comment "Did you forget an extra condition" – 2017-02-17
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    As in "Did you forget an extra condition that you want such a homomorphism to satisfy?" – 2017-02-17
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    I am happy to have no extra conditions just now, how can one prove that my above map is always true? – 2017-02-17
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    and one need to show that the map is epimorphism, how? – 2017-02-17
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    It is not always true that there exists a surjective map with this property. It is always true for finite abelian groups, but beyond that I am not sure that there is much you can say. – 2017-02-17

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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 $\mathfrak{T}$

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    such that the map is surjective (the map is epimorphism)? – 2017-02-17
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    I voted you +1, but I adjust my question earlier when you just posted the answer --- we need surjective -- so please add more statement – 2017-02-17
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    Sorry just saw your correction. I will edit my answer. – 2017-02-17
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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:

  1. $B\cap N=1$, and
  2. $A=BN$,

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.)