On Symmetric Group $S_n$ and Isomorphism

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 !

-
Did you read the proof? –  Vectk Jan 28 '13 at 20:48
The result you mention in your text is PROVEN on page $120$. Read the proof. If you don't understand the proof, ask for clarification. How to find an isomorphic group: "find a bijective function mapping group to group and such that it satifies the homomorphism property." –  amWhy Jan 28 '13 at 20:48
@YACP , yes , it wasn't mentioned explicitely how to do this in practise ! , go back to example in page 118 , you will find explanation why does V is isomorphic to this subgroup of Sn , but he didn't metion that this is an explanation , but when you finish reading the section and go back to read this example you will know that this is the explanation ! i think that the answers here made it obivous for me to how to calculate the subgroup of Sn which we look for it ! so , thanx for all ! –  Maths Lover Jan 28 '13 at 21:04

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.

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 $H$.
your answer is not what i look for ! why ? because i ask in general ! i don't ask about what is the subgroup generated by two elements ! i know this well !!! i ask , how can we find the subgroup H which V is isomorphic to it ? for example ! if G = $D_2n$ , then which subgroup H of $S_n% where G is isomorphic to H ??? how can we calculate H ??? how can we calculate H for any group G in general ???!!! that is my question ! – Maths Lover Jan 28 '13 at 20:35 In general? That's not what you asked. In general, find your answer in your text: p. 120. Corollary 4: Cayley's Theorem. Read the proof: that tells you in general WHY this is the case. I was just trying to answer your question. That is, to HELP. There is no need for exclamation points or anger; if you failed to ask the question you MEANT to ask, that is your lacking, not mine. I apologize if I've wasted both our time. – amWhy Jan 28 '13 at 20:42 @YACP , talking about klein group was an example from me to make what i mean obivous ! not as a question ! – Maths Lover Jan 28 '13 at 20:43 Maths Lover: Your question, I quote: "Now, if we have Klein 4-group, how can we know that Klein 4-group is isomorphic to H ? How can we calculate H ? " – amWhy Jan 28 '13 at 20:44 @MathsLover: You're welcome, but please see that the others here tried to clarify what are you looking for just by using$V$. ;-) – B. S. Jan 28 '13 at 20:47 show 5 more comments 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: 1.$(0, 0)$2.$(0, 1)$3.$(1, 0)$4.$(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.