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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

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    $\begingroup$ Do you know about composition of homomorphisms? $\endgroup$ Aug 20 '17 at 17:13
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    $\begingroup$ So, by Cayley's theorem, $G$ is isomorphic to a subgroup of $S_n$, and you know $S_n$ is isomorphic to a subgroup of ${\rm GL}_n(\Bbb C)$. You're unable to put those two facts together? $\endgroup$
    – anon
    Aug 20 '17 at 17:14
  • $\begingroup$ @JohnHughes I know that if $\varphi : G \rightarrow H$ is a homomorphism and $\theta : H \rightarrow K$ is a homomorphism then the composition of both is a homomorphism too. Is that what you mean? Also, is $S_n$ isomorphic to a subgroup of $GL_n(\mathbb{R})$? $\endgroup$
    – M.Byrne
    Aug 20 '17 at 17:18
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    $\begingroup$ Trying to spell it out. You have an injective homomorphism $f:G\to S_n$ by Cayley. You have an injective homomorphism $\phi:S_n\to GL_n(\Bbb{R})$. What can you say about $\phi\circ f$? $\endgroup$ Aug 20 '17 at 17:19
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    $\begingroup$ It doesn't need to be surjective, any map is surjective onto its image. $\endgroup$ Aug 20 '17 at 17:56
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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, $\text{Im}(\theta)$.

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  • $\begingroup$ I assume you are correct, but I have only understood a bit of your answer as out notes do not extend that far. $\endgroup$
    – M.Byrne
    Aug 21 '17 at 9:29
  • $\begingroup$ What do you want me to explain more ? $\endgroup$
    – Wirius
    Aug 21 '17 at 9:31
  • $\begingroup$ Well, we were meant to use the following information: If $f$ and $g$ are isomorphisms, then $g \circ f$ is an isomorphism, If $\ker (\varphi) = \{e\}$ then $\varphi$ is injective, an isomorphism is a bijective homomorphism and $\ker (\varphi)$ is a subgroup of $G$ if $\varphi : G \rightarrow H$ is a homomorphism. Also Cayley's theorem for finite groups. With that information we are meant to answer the question. $\endgroup$
    – M.Byrne
    Aug 21 '17 at 13:03
  • $\begingroup$ Ok, I understand what blocks you : I've only used injective morphisms, where you ask for a bijective morphism, is that right ? In that case, I'm going to change a bit of notations from my post above. Let $\varphi$ be an injective morphism from $G$ to $H$, where $G$ and $H$ are any groups. Then, we can define : $$\varphi_{\text{bij}} : \begin{matrix} G & \longrightarrow & \text{Im}(\varphi) \\ g & \longmapsto & \varphi(g) \end{matrix}$$ $\varphi_{\text{bij}}$ is an injective and surjective morphism, so $\varphi_{\text{bij}}$ is an isomorphism. $\endgroup$
    – Wirius
    Aug 21 '17 at 14:21
  • $\begingroup$ Let's apply this : by cayley, we have $\phi : G \to S_n$ which is injective, so by our process, we have $\phi_{\text{bij}}$ from $G$ to $\text{Im}(\phi)$ which is an isomorphism. In the same way as above, we define : $$\phi' : \begin{matrix} \text{Im}(\phi) & \longrightarrow & \text{GL}_n(\mathbb{R}) \sigma & \longmapsto & M_{\sigma} \end{matrix}$$ Let's study $\phi'$'s injectivity : $$\sigma \in \text{Ker}(\phi') \iff M_\sigma = I_n \iff \forall i \in [\![1, n]\!], \sigma(i) = i \iff \sigma = \iota$$ From that, we conclude that $\text{Ker}(\phi') = \{\iota\}$. $\endgroup$
    – Wirius
    Aug 21 '17 at 14:27
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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.

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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.

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    $\begingroup$ Of course that $\operatorname{im}(\varphi)\neq GL_n(\mathbb{R})$; $S_n$ is finite; whereas $GL_n(\mathbb{R})$ isn't. But you proved correctly that $GL_n(\mathbb{R})$ contains a subgroup isomorphic with $S_n$. $\endgroup$ Aug 20 '17 at 18:13

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