Seems all the proofs I saw are by construction, what if the construction is so hard that one can not possibly construct it by hand. Is it possible to prove the homomorphism without having to construct it? That's why I have this question.
(The tag should be homomorphism but I can only choose an existing one)
Generally speaking, either the construction will be relatively simple and intuitive or the objects won't be isomorphic at all. There is a way to avoid explicit constructions in certain cases, though: that's what invariants are for. Informally speaking, an invariant is a "juice" that you squeeze out of an object that depends only on a certain structure of that object. If you squeeze different juices out of different objects, then the objects can't be the same (as in, equivalent under the notion of equivalence you're working with - isomorphic, homeomorphic, diffeomorphic, ...). On the other hand, getting the same "juice" from different objects doesn't always guarantee the objects are equivalent. This is wonderful to decide when two objects can be equivalent. What are explicit examples of this?
The dimension of a finite vector space is an algebraic invariant: it completely determines (up to isomorphism) its vector space structure, i.e, finite vector spaces with the same dimension are isomorphic. The Gaussian/sectional curvature is a geometric invariant (depends only on the geometrical structure of the manifold): objects (in that case manifolds) with different sectional curvatures are never isometric. In this case the curvature doesn't determine the geometric structure, though: there are surfaces with identical Gaussian curvatures which are not locally isometric. On the other hand, the Gaussian curvature is a geometrical invariant of surfaces of constant Gaussian curvature: two surfaces with the same constant Gaussian curvature are necessarily isometric. In this case, the "juice" you extracted was the curvature. Other examples: the fundamental group is a topological invariant - an homeomorphism must always preserve the fundamental group. But spaces with the same fundamental group need not be even homotopy equivalent. The Euler characteristic is a topological invariant that doesn't determine the topology of your space, but it is a topological invariant of compact orientable (or non orientable) surfaces which does determine topology in that case.
Finding algebraic invariants for even finite groups is pretty hard though, it's actually a research topic.