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
Best Answer
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, $\text{Im}(\theta)$.