Show that any finite group $G$ is isomorphic to a subgroup of $GL_{n}(\mathbb{R})$ for some $n$.

Question

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

Answer 1

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.

Answer 2

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

Answer 3

This answer is similar to @Wirius's, but includes an example.

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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:

$$M_{ij} = (\phi(\sigma))_{ij} = \delta_{i \, \sigma(j)}$$

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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.