tauon said:you can use a certain (obvious) implication of the Lagrange theorem for groups to thin out the workload; and you can also use the fact that any permutation is a product of disjoint transpositions to further simplify notation... of course if by S4 you mean [tex]S_4[/tex] the symmetry group.
so yes there is a way to find the normal subgroups by other means than brute force.
tauon said:you can use a certain (obvious) implication of the Lagrange theorem for groups to thin out the workload; and you can also use the fact that any permutation is a product of disjoint transpositions to further simplify notation... of course if by S4 you mean [tex]S_4[/tex] the symmetry group.
so yes there is a way to find the normal subgroups by other means than brute force.
A normal subgroup of S4 is a subgroup of the symmetric group S4 that is invariant under conjugation. This means that when you apply a permutation to the elements of the subgroup, the result is still within the subgroup.
To determine if a subgroup of S4 is normal, you can use the normal subgroup test. This involves checking if the subgroup is closed under conjugation by all elements of S4. If the subgroup remains unchanged after conjugation, it is normal.
The order of a normal subgroup of S4 is a divisor of the order of S4, which is 24. This means that the order of a normal subgroup can be 1, 2, 3, 4, 6, 8, 12, or 24.
Yes, a normal subgroup of S4 can be a cyclic group. For example, the subgroup generated by the permutation (1 2 3) is a cyclic subgroup of S4 and it is also normal.
The normalizer of a subgroup in S4 is the largest subgroup of S4 that contains the subgroup and is normal. This means that the normalizer of a subgroup is also a normal subgroup of S4. However, not all normal subgroups are normalizers, as there may be larger normal subgroups that contain the given subgroup.