For the definition of see here. The group of permutations of is denoted by

**Definition**. Let and let be the standard ordered basis of as a vector space over Let and define the linear transformation by for all Then the matrix of with respect to is denoted by and is called a **permutation matrix**.

**Example**. Let Then and Choose Then is defined by

Thus

**Remark 1**. If then In particular, the inverse of is for all

*Proof*.

**Problem**. Prove that every finite group of order is isomorphic to a subgroup of

**Solution**. By Cayley’s theorem a finite group of order is isomorphic to a subgroup of So we only need to prove that is isomorphic to a subgroup of So we need to define an injective group homomorphism from to Define by

1) is well-defined. Because clearly and, by Remark 1, Thus and so is well-defined.

2) is a group homomorphism. Because, by Remark 1 again, for all and so is a group homomorphism.

3) is injective. Because if then is the identity matrix. Thus is the identity linear transformation, i.e. for all But, by definition, and so for all Hence for all i.e. is the identity permutation.

**Remark 2**. In here we proved that every finite subgroup of is isomorphic to a subgroup of for any prime number So, by the above problem, every finite group of order is isomorpic to a subgroup of for any prime number

Oh i get i now. Thanks

Hi —

I have a doubt. You have shown that is a isomorphism, which says that . But your claim was to show that S_n is isomorphic to a subgroup of GL(n,Z).

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? :))