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.

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Comments
  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? :))

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