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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 $G$ ?

- Enumerating all subgroups of the symmetric group
- Definition of the Young Symmetrizer
- Disjoint Cycles Cannot Be Inverses
- Show that $G/H\cong S_3$
- Splitting of conjugacy class in alternating group
- Proving that any permutation in $S_n$ can be written as a product of disjoint cycles

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 !

- Probability that two randomly chosen permutations will generate $S_n$.
- $A_4$ has no subgroup of order $6$?
- Confusing partitions of $S_5$ in two different sources
- Point Group of a pattern
- Normal subgroups of the symmetric group $S_N$
- Generating the symmetric group $S_n$
- How to enumerate subgroups of each order of $S_4$ by hand
- Show that any abelian transitive subgroup of $S_n$ has order $n$

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:

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

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.

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