How to write the set of all permutations on a set $n=\{1, 2, \ldots, n\}$ Let $n ∈ N$. Let $S_n$ denote the set of permutations on $\{1, . . . , n\}$. 
For any $σ ∈ S_n$, define $sign(σ) := (−1)^N$ , where $σ$ can be written as the product of $N$ transpositions.
Now, let $A$ be an $n \times n$ matrix with entries $A_{ij},i,j∈\{1,...,n\}$. Consider the expression:


*

*$F(A) := \sum_{σ ∈ S_n}sign(σ) \prod_{i=1}^{n} A_{iσ(i)}.$


QUESTION:
If we have a $2 \times 2$ matrix $A$, we have $n =2$. I want to know how the set $S_n$ would look like.
Thank you.

Say you have the matrix 
$$A= \begin{pmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{pmatrix},$$
a very normal $2\times 2$ matrix. 
As you said $S_n$ is the set of all the permutations of the set $\{1,\dots,n\}$. So $S_2$ is the set of all the permutations of $\{1,2\}$, hence
$$S_2 = \{[12],[21]\}.$$
You can think at the function $\sigma$ as follows: $\sigma(i)=j$ means the permutation $\sigma$ sends the number $i$ to $j$. 
The permutation $\sigma_1:=[12]$ is the identity of the set $S_2$, since the number $1$ and the number $2$ do not change their position. Hence we have $\sigma_1(1)=1$ and $\sigma_1(2)=2$. The permutation $\sigma_2:=[21]$ actually says that $1$ goes to $2$ and that $2$ goes to $1$, so $\sigma_2(1)=2$ and $\sigma_2(2)=1$. These two are the only possible permutations of the set $\{1,2\}$. This is how $S_2$ looks like.
We can also look at how $$F(A) := \sum_{σ ∈ S_2}\text{sign}(σ) \prod_{i=1}^{2} A_{iσ(i)}$$ looks like. First of all the term $\sum_{σ ∈ S_2}$ tells you that you need to sum $2$ elements, since you have only $2$ permutations in $S_2$. As a second step you want to evaluate the signum of the permutations in $S_2$. You defined it as $\text{sign}(\sigma) = (-1)^N$, where $N$ is the number of transpositions of your permutation. We have two signum to evaluate:
$$\text{sign}(\sigma_1)= (-1)^0=1; \qquad \text{sign}(\sigma_2) = (-1)^1=-1.$$
(ask if you have problem with this step). Now we can finally evaluate $F(A)$:
\begin{align*}
F(A) &= \sum_{σ ∈ S_2}\text{sign}(σ) \prod_{j=1}^{2} A_{jσ(j)} \\
&= \sum_{i=1}^2 \text{sign}(\sigma_i) \prod_{j=1}^2 A_{j\sigma_i(j)} \\
&= \text{sign}(\sigma_1)(A_{1\sigma_1(1)}A_{2\sigma_1(2)}) + \text{sign}(\sigma_2)(A_{1\sigma_2(1)}A_{2\sigma_2(2)}) \\
&= A_{11}A_{22}-A_{12}A_{21}.
\end{align*}
This is actually the determinant of the matrix $A$. Indeed $F(A)$, in general, is exactly the determinant of an $n\times n$ matrix. Don't esitate to ask if you have problems.
