## Embedding of finite groups into GL(n,Z)

Posted: February 18, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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For the definition of $\text{GL}(n,\mathbb{Z})$ see here. The group of permutations of $\{1,2, \cdots , n \}$ is denoted by $S_n.$

Definition. Let $n \geq 1$ and let $\mathcal{B}=\{e_1, \cdots , e_n\}$ be the standard ordered basis of $\mathbb{R}^n,$ as a vector space over $\mathbb{R}.$ Let $\sigma \in S_n$ and define the linear transformation $T_{\sigma}: \mathbb{R}^n \longrightarrow \mathbb{R}^n$ by $T_{\sigma}(e_i) = e_{\sigma(i)},$ for all $i=1, \cdots , n.$ Then $[T_{\sigma}]_{\mathcal{B}},$ the matrix of $T_{\sigma}$ with respect to $\mathcal{B},$ is denoted by $P_{\sigma}$ and is called a permutation matrix.

Example. Let $n=3.$ Then $e_1=(1,0,0), \ e_2=(0,1,0)$ and $e_3=(0,0,1).$ Choose $\sigma = (1 \ \ 3 \ \ 2) \in S_3.$ Then $T_{\sigma} : \mathbb{R}^3 \longrightarrow \mathbb{R}^3$ is defined by

$T_{\sigma}(e_1)=e_{\sigma(1)}=e_3, \ T_{\sigma}(e_2)=e_{\sigma(2)}=e_1, \ T_{\sigma}(e_3)=e_{\sigma(3)}=e_2.$

Thus

$P_{\sigma} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}.$

Remark 1. If $\sigma_1, \sigma_2 \in S_n,$ then $P_{\sigma_1 \sigma_2}=P_{\sigma_1}P_{\sigma_2}.$ In particular, the inverse of $P_{\sigma}$ is $P_{\sigma^{-1}},$ for all $\sigma \in S_n.$

Proof. $T_{\sigma_1}T_{\sigma_2}(e_i)=T_{\sigma_1}(e_{\sigma_2(i)})=T(e_{\sigma_1 \sigma_2(i)})=T_{\sigma_1 \sigma_2}(e_i). \Box$

Problem. Prove that every finite group of order $n$ is isomorphic to a subgroup of $\text{GL}(n,\mathbb{Z}).$

Solution. By Cayley’s theorem a finite group of order $n$ is isomorphic to a subgroup of $S_n.$ So we only need to prove that $S_n$ is isomorphic to a subgroup of $\text{GL}(n,\mathbb{Z}).$ So we need to define an injective group homomorphism from $S_n$ to $\text{GL}(n,\mathbb{Z}).$ Define $\varphi : S_n \longrightarrow \text{GL}(n,\mathbb{Z})$ by

$\varphi(\sigma)=P_{\sigma}.$

1) $\varphi$ is well-defined. Because clearly $P_{\sigma} \in M_n(\mathbb{Z})$ and, by Remark 1, $(P_{\sigma})^{-1}=P_{\sigma^{-1}} \in M_n(\mathbb{Z}).$ Thus $P_{\sigma} \in \text{GL}(n,\mathbb{Z})$ and so $\varphi$ is well-defined.

2) $\varphi$ is a group homomorphism. Because, by Remark 1 again, $\varphi(\sigma_1 \sigma_2)=P_{\sigma_1 \sigma_2}=P_{\sigma_1}P_{\sigma_2}=\varphi(\sigma_1) \varphi(\sigma_2),$ for all $\sigma_1, \sigma_2 \in S_n$ and so $\varphi$ is a group homomorphism.

3) $\varphi$ is injective. Because if $\sigma \in \ker \varphi,$ then $\varphi(\sigma)=P_{\sigma}$ is the identity matrix. Thus $T_{\sigma}$ is the identity linear transformation, i.e. $T_{\sigma}(e_i)=e_i$ for all $i.$ But, by definition, $T_{\sigma}(e_i)=e_{\sigma(i)}$ and so $e_i=e_{\sigma(i)},$ for all $i.$ Hence $\sigma(i)=i,$ for all $i,$ i.e. $\sigma$ is the identity permutation. $\Box$

Remark 2. In here we proved that every finite subgroup of $\text{GL}(n,\mathbb{Z})$ is isomorphic to a subgroup of $\text{GL}(n, \mathbb{F}_q)$ for any prime number $q > 2.$ So, by the above problem, every finite group of order $n$ is isomorpic to a subgroup of $\text{GL}(n,\mathbb{F}_q)$ for any prime number $q > 2.$

1. Chandrasekhar says:

Oh i get i now. Thanks

2. Chandrasekhar says:

Hi —

I have a doubt. You have shown that $\phi: S_{n} \to GL(n,\mathbb{Z})$ is a isomorphism, which says that $S_{n} \cong GL(n,\mathbb{Z})$. But your claim was to show that S_n is isomorphic to a subgroup of GL(n,Z).

• Yaghoub says:

No! I only proved that the map is “injective” and so S_n is isomorphic to a “subgroup” of GL(n,Z). You should probably look at the proof again.
Besides, how can the map be an “isomorphism” when GL(n,Z) is infinite but S_n is finite? :))