# Group Theory/The symmetric group

**Definition (symmetric group)**:

Let be a set. Then the **symmetric group** of is defined to be

- ;

that is, it is the set of all bijective functions from to itself with composition as operation.

**Definition (permutation)**:

A **permutation** is, by definition, an element of .

**Proposition (symmetric group essentially depends only on the cardinality of the underlying set)**:

Let be sets of the same cardinality. Then there exists a group isomorphism

- .

**Proof:** Suppose that is a bijective function. Then the group isomorphism is given by

- ;

indeed, an inverse is given by

- .

**Definition (finite symmetric group)**:

Let . Then the **symmetric group of order **, denoted , is defined to be

- .

**Theorem (Cayley's theorem)**:

Let be a finite group, and set . Then there exists a subgroup of which is isomorphic to .

**Proof:** acts transitively on itself by left multiplication in the category of sets. This means that we have a group homomorphism . Moreover, this morphism is injective; indeed, only the identity element of induces the identity element in . Hence, the claim follows from the first Noether isomorphism theorem.

**Definition (matrix representation of a permutation)**:

The representation of in the category of vector spaces over a field given by

is called the **matrix representation** of the permutations contained within .

**Definition (sign)**:

Let be a permutation. Then the **sign** of , written , is defined to be , where there exist transpositions such that .

The following proposition shows that this notion is well-defined:

**Proposition (equivalent characterisations of the sign of a permutation)**:

**Definition (alternating group)**:

Let . Then the **alternating group**, a subgroup of , is defined to be

- .

**Proposition (alternating group is maximal and normal in the symmetric group)**:

Let . Then , and further is a maximal subgroup of .

In particular, is a maximal normal subgroup in (ie. maximal among the normal subgroups).

**Proof:** Note first that is normal as the kernel of a group homomorphism. We then have that is a group homomorphism from to , and by the first Noether isomorphism theorem, . In particular, there are only two cosets of . Suppose that there existed a subgroup . Then by the degree formula, we would have , so that either or . In both cases, one of the inclusions is not strict, a contradiction.

**Proposition (conjugation in the symmetric group is re-labeling)**:

Let be a cycle, and let be any element. Then

- .

**Proposition (in degrees 5 or larger all three-cycles are conjugate in the alternating group)**:

Let , and let and be any two three-cycles in . Then there exists such that .

**Proof:** Since being in the same conjugacy class is an equivalence relation, assume .

**Theorem (in degrees 5 or larger the alternating group is simple)**:

Let . Then is a simple group.

**Proposition (in degrees 5 or larger neither the alternating nor the symmetric group are solvable)**:

Let . Then and are both not solvable.

**Proof:** Since is a maximal normal subgroup of ,