Finite Groups, p Cycles and Normal Subgroups

p Cycles and Normal Subgroups

The main reason for this theorem is to prove that the group of order 168, described in a later section, is simple. However, the theorem can be applied to other groups as well. It is presented in small "pieces".

  1. Let G be a permutation group within Sn. Assume the action is transitive, hence there is one orbit of size n. The stabilizer of any element is a subgroup of index n. All these stabilizers are conjugate, since they lie in the same orbit.

    Suppose a stabilizer contains K, a normal subgroup of G. Conjugate the stabilizer so that it fixes another element of S. Thus the conjugate of K, which equals K, fixes the second element. And the third, and the fourth, etc, hence K is trivial. If G is transitive, K does not live within a stabilizer.

  2. Let G be transitive and let T be a block in S. Let H be the subgroup of G that maps the block T onto itself. Let J be the stabilizer of an element x in T. Since T is a block, J is wholly contained in H. Cosets of J move x to other elements in T, hence J has index |T| in H. Thus |T| divides the index of J in G, which is n.

    When n is prime, all blocks have size 1 or n, and G is primitive. For n prime, G is primitive iff G is transitive.

  3. Let G be a primitive permutation group on n letters. Suppose the stabilizer of x, call it J, is contained in a larger subgroup H. We will see that this cannot happen.

    Let T be the orbit of x with respect to H, so that H maps T onto itself. Since J is contained in H, each coset of J is entirely inside H or entirely outside H. The rest of G, outside of H, contains cosets of J that move x outside of T. This is true for every x in T, hence T is a block. Since G is primitive, T is the single element x or all of S. There is no subgroup strictly between J and G.

  4. Assume G is primitive on n letters, J is the stabilizer of x, and K is a normal subgroup of G. Recall that K is not contained in J (1), so K join J is properly larger than J, hence K join J = G (3). If R is K∩J, a dimensionality argument shows the index of R in K = the index of J in G = n. Therefore |K| is divisible by n.

  5. Let G be a primitive group on p letters, where p is prime. The order of G is p!, which has only one factor of p. Let K be a normal subgroup. By (4), p divides |K|. If x, outside of K, has order p, then the image of x in G/K has order p. Yet G/K is not divisible by p, so every element of order p lies in K. Every p cycle is contained in K. A normal subgroup contains all the p cycles.

  6. Let G be transitive on p letters, and let K be a normal subgroup. Since G is transitive it is primitive (2). A normal subgroup K contains all the p cycles of G (5).