TITLE: On Symmetric Group $S_n$ and Isomorphism
QUESTION [7 upvotes]: 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 !

REPLY [4 votes]: 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:

$(0, 0)$
$(0, 1)$
$(1, 0)$
$(1, 1)$

Then left 'multiplication' (actually addition in this case) by $(1, 0)$ sends

$(0, 0) \to (1, 0)$
$(0, 1) \to (1, 1)$
$(1, 0) \to (0, 0)$
$(1, 1) \to (0, 1)$

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.