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 ,