Alternating Groups #
The alternating group on a finite type α is the subgroup of the permutation group Perm α
consisting of the even permutations.
Main definitions #
alternatingGroup αis the alternating group onα, defined as aSubgroup (Perm α).
Main results #
alternatingGroup.index_eq_twoshows that the index of the alternating group is two.two_mul_card_alternatingGroupshows that the alternating group is half as large as the permutation group it is a subgroup of.closure_three_cycles_eq_alternatingshows that the alternating group is generated by 3-cycles.alternatingGroup.isSimpleGroup_fiveshows that the alternating group onFin 5is simple. The proof shows that the normal closure of any non-identity element of this group contains a 3-cycle.Equiv.Perm.eq_alternatingGroup_of_index_eq_twoshows that a subgroup of index 2 ofEquiv.Perm αis the alternating group.Equiv.Perm.alternatingGroup_le_of_index_le_twoshows that a subgroup of index at most 2 ofEquiv.Perm αcontains the alternating group.
Instances #
- The alternating group is a characteristic subgroup of the permutaiton group.
Tags #
alternating group permutation simple characteristic index
TODO #
- Show that
alternatingGroup αis simple if and only ifFintype.card α ≠ 4.
The alternating group on a finite type, realized as a subgroup of Equiv.Perm.
For $A_n$, use alternatingGroup (Fin n).
Equations
Instances For
Equations
Equations
- instUniqueSubtypePermMemSubgroupAlternatingGroupOfSubsingleton α = { default := 1, uniq := ⋯ }
A key lemma to prove $A_5$ is simple. Shows that any normal subgroup of an alternating group on at least 5 elements is the entire alternating group if it contains a 3-cycle.
Part of proving $A_5$ is simple. Shows that the square of any element of $A_5$ with a 3-cycle in its cycle decomposition is a 3-cycle, so the normal closure of the original element must be $A_5$.
The normal closure of $(04)(13)$ within $A_5$ is the whole group. This will be used to show that the normal closure of any permutation of cycle type $(2,2)$ is the whole group.
Shows that any non-identity element of $A_5$ whose cycle decomposition consists only of swaps is conjugate to $(04)(13)$. This is used to show that the normal closure of such a permutation in $A_5$ is $A_5$.
Shows that $A_5$ is simple by taking an arbitrary non-identity element and showing by casework on its cycle type that its normal closure is all of $A_5$.
The alternating group is the only subgroup of index 2 of the permutation group.
A subgroup of the permutation group of index ≤ 2 contains the alternating group.
The alternating group is a characteristic subgroup of the permutation group.