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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 !

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Did you read the proof? – Ink 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
up vote 4 down vote accepted

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 $H$.

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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 $V$. ;-) – Babak S. Jan 28 '13 at 20:47

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.

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