Show that any finite group $G$ is isomorphic to a subgroup of $GL_{n}(\mathbb{R})$ for some $n$. We are told the following:
Let $S_n$ denote the permutation group on $\{1,\dots,n\}$ and let $GL_n(\mathbb{R})$ denote the group of invertible $n \times n$ matrices.  Now assume the following fact:
for each $n$ there is a group homomorphism $\varphi : S_n \rightarrow GL_n(\mathbb{R})$ such that $\ker (\varphi) = \{\iota\}$, where $\iota$ is the identity permutation.
How do I show that any finite group $G$ is isomorphic to a subgroup of $GL_{n}(\mathbb{R})$ for some $n$?
I realise I need to use Cayley's theorem for a finite group which is:
Every finite group $G$ of order $n$ is isomorphic to a subgroup of $S_n$.
However, I am unsure how I can answer this question. 
Thank you
 A: You've almost made it to the end !
Let $G$ be a finite groupe and $n = \text{Card}(G)$.
Caley's theorem states that there exists $\phi \in \text{Hom}(G, S_n)$ which is injective, which means that $G$ is isomorphic to a subgroup of $S_n$.
We will make $S_n$ "act" on the vector space $\mathbb{R}^n$. We will define for every $\sigma \in S_n$ the matrix :
$$M_{\sigma} = (\delta_{\sigma(i), j})$$
where $\delta$ is the Kronecker delta defined by : $\delta_{a, b} = \begin{cases}
1 \text{ if } a = b \\
0 \text{ otherwise}
\end{cases}$
These $M_{\sigma}$ are elements of $\text{GL}_n(\mathbb{R})$ because $\forall \sigma \in S_n, \ M_{\sigma} \times M_{\sigma^{-1}} = M_{\sigma^{-1}} \times M_{\sigma} = I_n$.
Let $\phi' : \sigma \mapsto M_{\sigma}$. $\phi' \in \text{Hom}(S_n, \text{GL}_n(\mathbb{R}))$ and is injective, so $S_n$ is isomorphic to a subgroup of $\text{GL}_n(\mathbb{R})$.
then, let $\theta = \phi' \circ \phi$. $\phi$ and $\phi'$ are injective morphisms, so $\theta$ is an injective morphism from $G$ to $\text{GL}_n(\mathbb{R})$. That means $G$ is isomorphic to a subgroup of $\text{GL}_n(\mathbb{R})$, which is, in our case, $\text{Im}(\theta)$.
A: This answer is similar to @Wirius's, but includes an example.

We seek a homomorphism $\phi : S_n \rightarrow \text{GL}_n (\mathbb{R})$.

*

*Let $\sigma \in S_n$ be a permutation that sends $[1, n] \rightarrow [1, n]$.

*Let $M \in \text{GL}_n (\mathbb{R})$ be a matrix that sends a vector $\mathbb{R}^n \rightarrow \mathbb{R}^n$.

Then one simple identification is:
$$M_{ij} = (\phi(\sigma))_{ij} = \delta_{i \, \sigma(j)}$$

To see why this is a good choice, let $n = 3$ and $\sigma = (1 \; 3 \; 2)$. Then:
$$\phi(\sigma) = \begin{pmatrix}
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    1 & 0 & 0
\end{pmatrix}$$
This matrix acts on a vector $\vec{x} = (x_1 \; x_2 \; x_3)^T$ to give:
$$\begin{pmatrix}
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
    x_1 \\ x_2 \\ x_3
\end{pmatrix}
=
\begin{pmatrix}
    x_2 \\ x_3 \\ x_1
\end{pmatrix}$$
Just as we expect.
A: By Cayley's theorem for finite groups:
Every finite group $G$ of order $n$ is isomorphic to a subgroup of $S_n$.
Let $f: G \rightarrow S_n$ be a homomorphism and also an isomorphism.
We also know that $\varphi : S_n \rightarrow GL_n(\mathbb{R})$ is a homomorphism.  
Since $\ker (\varphi) = \{\iota\}$ where $\iota$ is the identity permutation, then $\varphi$ is injective.  Furthermore since $\text{im} (\varphi) = GL_n(\mathbb{R})$ then $\varphi$ is also surjective.  This means that $\varphi$ is a bijective homomorphism, which means it is an isomorphism.  
Therefore since $f$ and $\varphi$ are both isomorphisms, then $\varphi \circ f$ is an isomorphism too.  Hence any finite group $G$ is isomorphic to a subgroup of $GL_n(\mathbb{R})$ for some $n$.
This would be my attempt at the answer.  
