Group Actions, Strong Cayley Theorem

Strong Cayley Theorem

Any group G with subgroup H acts on left cosets of H by left multiplication. Remember that the action of ab is b first, then a, so bH creates a left coset, then abH gives another left coset. This is a proper group action on a set whose size is the index of H in G.

Let f be the homomorphism defined by this action. The quotient group is a permutation group on n letters, where n is the index of H. The kernel is the set of elements a fixing every coset. Let x be any cosrep, hence a*x is in the same coset as x. that is, a*x = x*j for some j in H. Multiply by x inverse on the right and a = xj/x. Thus the kernel is the intersection of all the conjugates of the subgroup H. We know this is a normal subgroup, because it's the kernel of the action of G. Call this normal subgroup K. Thus G/K embeds in Sn, where n is the index of H in G.

The trivial conjugate of H is H, hence K is a subgroup of H. In fact K is the largest normal subgroup of G contained in H. If some other normal subgroup M is contained in H, x*M/x = M for all x, and the intersection of x*H/x would include all of M.

Constraints on a Subgroup of Prime Index

Let H be a subgroup of G with index p, where p is prime. Let K be the largest normal subgroup of G contained in H. By the strong Cayley theorem, K is the kernel of the homomorphism induced by left translation on the cosets of H. Thus G/K embeds in Sp, and the order of G/K divides p!.

Let q be a prime ≥ p, such that q divides |H|. Now q either divides the order of K or the index of K in H. If the latter then qp, the index of K in H times the index of H in G, divides p!, which is impossible. Hence q, and every higher power of q for that matter, must divide |K|.

If we know that every prime q in |H| is at least as large as p, then K = H and H is normal. This is a generalization of an earlier result; that H is normal in G if its index is 2.