How to convert a permutation group into linear transformation matrix? is there any example about apply isomorphism to permutation group
and how to convert linear transformation matrix to permutation group and convert back to linear transformation matrix
 A: It seems that for each natural number $n$ there is an isomorphic embedding $i$ of the permutation group $S_n$ into the group of all non-degerated matrices of order $n$ over $\mathbb R$, defined as $i(\sigma)=A_\sigma=\|a_{ij}\|$ for each $\sigma\in S_n$, where $a_{ij}=1$ provided $\sigma(i)=j$, and $a_{ij}=0$ in the opposite case.
From the other side, if $G$ is any group (and, in particular, a matrix group of linear transormations) then any element $g\in G$ induces a permutation $j(g)$ of the set $G$ such that $j(g)h=gh$ for each $h\in G$. Then the map $j:G\to S(G)$ should be an isomorphic embedding of the group $G$ into the group $S(G)$ of all permutations of the set $G$. You can read more details about such embedding here.
A: Hint:Prove that Permutatation matrices form a subgroup of $\mathbb{GL}_n(R)$(under multiplication) and this subgroup is isomorphic to $S_n$.
You can visualize it if you think of the way permutation matrices work when they are applied from the left to any matrix. 
Permutation matrices just change the positions of some rows when they are applied from the left to some matrix.
Multiplying two permutation matrix results in another permutation matrix because In a matrix if we apply two permutation matrices from the left we are left with another matrix haing the same rows but in a different position so it can again be represented by another permutation matrix.
Clearly the permutation matrix doing the reverse change of these rows will be its inverse.
The homomorphism that shows the two groups to be isomorphic is $\phi(P)=(\sigma_1,\sigma_2,\dots ,\sigma_n)$. Here $\sigma_i $ is the row number where the $i$ th row of any matrix go when $P$ is applied to it from the left.
A: $$\varphi:\begin{array}{ll}\mathcal S_n \to \mathcal L\left(\Bbb R^n\right)\\ \sigma \mapsto \left(x = x_1b_1 + \dots + x_n b_n \mapsto x_1b_{\sigma(1)} + \dots + x_n b_{\sigma(n)} \right)\end{array} $$
You can prove it is an injective morphism so if you restrict its codomain to its image, you get and isomorphism.
And then you just need to have an isomorphism from $\mathcal L\left(\Bbb R^n\right)$ to $\mathcal M_n\left(\Bbb R\right)$ to get the isomorphism you want.
