I use Abstract Algebra by Dummit and Foote to study abstract algebra! At page 120, section 2 in chapter 4, there is a great result form my point of view which proves that, for any group $G$ of order $n$, $G$ is isomorphic to some subgroup of $S_n$.
My question: Is there any way to calculate the subgroup of $S_n$ which is isomorphic to some group $G$ ?
I mean, if we have a group $G$, how can we calculate the subgroup of $S_n$ which $G$ is isomorphic to it ? my question is in general !
the question is edited !
You find a bijective map $\varphi$ from the klein 4-group to $H$ such that $\varphi$ satisfies the homorphism property:
For example: Let $V$ denote the Klein 4-group. Then you find a bijective function mapping identity to identity, with $\phi: V \to H$ such that
$$\forall a, b \in V, \;\varphi(ab) = \varphi(a)\circ\varphi(b)$$
where $ab$ denotes the product operation of $V$, and $\circ$ denotes permutation composition.
In answer to your original question…
To find the subgroup of $S_n$ generated by $(12)(34)$ and $(13)(24)$, take products and inverses to obtain closure. There will be the identity, each of these elements, and the product of these elements, which will be $(14)(23)$. Each element is its own inverse. So we have a subgroup $H \leq S_4$ of order $4$.
Using the above procedure, you should be able to construct an isomorphism by the proper assignment of elements of $V$ to elements of $H$.
The proof of that theorem tells you exactly how to find the subgroup you are looking for. You number the elements of your group from $1$ to $n$. Left multiplication by an element of your group then corresponds to a permutation of these numbers.
For example the Klien $4$-group is $\mathbb Z/2 \times \mathbb Z/2$ so we number the elements as such:
Then left ‘multiplication’ (actually addition in this case) by $(1, 0)$ sends
Hence according to our numbering the element $(1, 0)$ is sent to the permutation $(1 \ 3)(2 \ 4)$. Likewise you can show that $(0, 1)$ is sent to $(1 \ 2)(3 \ 4)$ under this map.