normal subgroups of the symmetric groups

Theorem 1.

For n5, An is the only proper nontrivial normal subgroupMathworldPlanetmath of Sn.


This is essentially a corollary of the simplicity of the alternating groupsMathworldPlanetmath An for n5. Let NSn be normal. Clearly NAnAn. But An is simple, so NAn=An or NAn={e}. In the first case, either N=An, or else N also contains an odd ( permutationMathworldPlanetmath, in which case N=Sn. In the second case, either N={e} or else N consists solely of one or more odd permutationsMathworldPlanetmath in additionPlanetmathPlanetmath to {e}. But if N contains two distinct odd permutations, σ and τ, then either σ2e or στe, and both σ2 and στ are even (, contradicting the assumptionPlanetmathPlanetmath that N contains only odd nontrivial permutations. Thus N must be of order 2, consisting of a single odd permutation of order 2 together with the identityPlanetmathPlanetmathPlanetmath.

It is easy to see, however, that such a subgroupMathworldPlanetmathPlanetmath cannot be normal. An odd permutation of order 2, σ, has as its cycle decomposition one or more (an odd number, in fact, though this does not matter here) of disjoint transpositionsMathworldPlanetmath. Suppose wlog that (12) is one of these transpositions. Then τ=(13)σ(13)=(13)(12)()(13) takes 2 to 3 and thus is neither σ nor e. So this group is not normal. ∎

If n=1, S1 is the trivial group, so it has no nontrivial [normal] subgroups.

If n=2, S2=C2, the unique group on 2 elements, so it has no nontrivial [normal] subgroups.

If n=3, S3 has one nontrivial proper normal subgroup, namely the group generated by (123).

S4 is the most interesting case for n5. The arguments in the theorem above do not apply since A4 is not simple. Recall that a normal subgroup must be a union of conjugacy classesMathworldPlanetmathPlanetmath of elements, and that conjugate elements in Sn have the same cycle type. If we examine the sizes of the various conjugacy classes of S4, we get

Cycle Type Size
4 6
3,1 8
2,2 3
2,1,1 6
1,1,1,1 1

A subgroup of S4 must be of order 1,2,3,4,6,8, or 12 (the factors of |S4|=24). Since each subgroup must contain {e}, it is easy to see that the only possible nontrivial normal subgroups have orders 4 and 12. The order 4 subgroup is H={e,(12)(34),(13)(24),(14)(23)}, while the order 12 subgroup is A4. A4 is obviously normal, being of index 2, and one can easily check that HV4 is also normal in S4. So these are the only two nontrivial proper normal subgroups of S4.

Title normal subgroups of the symmetric groupsMathworldPlanetmathPlanetmath
Canonical name NormalSubgroupsOfTheSymmetricGroups
Date of creation 2013-03-22 17:31:38
Last modified on 2013-03-22 17:31:38
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 8
Author rm50 (10146)
Entry type Theorem
Classification msc 20B35
Classification msc 20E07
Classification msc 20B30