Construction of an injective homomorphism from $S_n$ to $GL(n,\mathbb R)$ 
I am trying to define an injective homomorphism from $S_n$ to $GL(n,\mathbb R)$.

I simply don't have any idea how to start with. Any hint or suggestion will be appreciated.
If I define $f:S_n→GL(n,\mathbb R)$ by $f(σ)=A=(A_{ij})$, where $$A_{ij} = \left\{ \begin{array}{ll} 1  & \mbox{if } \sigma(j)=i \\ 0 & \mbox{if } \sigma(j)\ne i \end{array}\right.$$ will it work?
But the problem is how to show this is a homomorphism? Please help.
 A: Hint: Let $S_n$ act on the $n$ element set given by the standard basis of $\mathbb R^n$.
A: With your definition: $(f(\sigma \cdot \tau))_{ij}=\begin{cases}1& : (\sigma \cdot \tau)(j)=i\\
0& : otherwise \end{cases}$ on the other hand by the matrix product formula $$(f(\sigma)f(\tau))_{ij}=\sum\limits_{k=1}^n(f(\sigma))_{ik}(f(\tau))_{kj} =\begin{cases}1& : \tau(j)=k \;\;and\;\;\sigma(k)=i\\
0& : otherwise \end{cases}$$ becuase $(f(\sigma))_{ik}$ and $(f(\tau))_{kj}$ are non null iff $\sigma(k)=i$ and $\tau(j)=k$.
Notice that  the condition $\tau(j)=k \;\;and\;\;\sigma(k)=i$ is equivalent to $i=\sigma(\tau(j))=(\sigma\cdot\tau )(j)$ thus proving that $f(\sigma\cdot\tau)=f(\sigma)f(\tau)$ hence $f$ is a homeomorphism.
With a simmilar idea you should be able to prove that $f(\sigma)=Id_n$ iff $\sigma=Id\in S_n$.
A: Define the homomorphism as f(g) = Ag where Ag is the identity matrix with its rows interchanged. For example, if the permutation is (12)(34) in S4, the corresponding matrix would be the identity matrix with row 1 and 2 interchanged and rows 3 and 4 interchanged
A: Remember from Representation theory that,
$\text{A representation of a group $G$ on a linear space $V$ is the same as a group homomorphism}$ $\text{ from $G$ to $GL(V)$}.$
Thus define group action of $S_n$ on $\mathbb{R}^n$, which will give you the homormorphism 
A: Any action of a group  $G $ on a set $X $ gives a homomorphism from  $G $ to ${\rm Sym}(X) $.
So considering the action on the standard basis $X=\{e_1,e_2,\dots,e_n\} $ of $\mathbb R^n $,  it suffices to check that ${\rm Sym}(X)\le{\rm GL}(n,\mathbb R) $.
Indeed,  these correspond to  the so-called permutation matrices, with one $1$ in each row and column,  $0$'s everywhere else. Just take  $\sigma\in S_n $ to the transformation whose  matrix  has  $i $-th column equal to $e_{\sigma (i)} $.
