Group Theory/The symmetric group

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. permutation A permutation is, by definition, an element of . 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 . Suppose that is a bijective function. Then the group isomorphism is given by ; indeed, an inverse is given by . finite symmetric group Let . Then the symmetric group of order , denoted , is defined to be . Cayley's theorem Let be a finite group, and set . Then there exists a subgroup of which is isomorphic to . 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. 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 . 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: equivalent characterisations of the sign of a permutation alternating group Let . Then the alternating group, a subgroup of , is defined to be . 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). 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. conjugation in the symmetric group is re-labeling Let be a cycle, and let be any element. Then . 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 . Since being in the same conjugacy class is an equivalence relation, assume . in degrees 5 or larger the alternating group is simple Let . Then is a simple group. in degrees 5 or larger neither the alternating nor the symmetric group are solvable Let . Then and are both not solvable. Since is a maximal normal subgroup of ,