Group Actions, Second Sylow Theorem

Second Sylow Theorem

All sylow subgroups are conjugate.

Let G have order mpⁿ, where p does not divide m. Let H and J be any two sylow subgroups of G. Thus |H| = |J| = pⁿ.

Let H act on the left cosets of J by left translation. Recall the reasoning used in the congruent index principle. The coset xJ is fixed by H iff yxJ = xJ for all y in H, iff the conjugate of y maps J onto J, iff the conjugate of y is in J, iff xH/x lies in J. When |H| = |J|, this means H and J are conjugate through x. Thus a coset of J, represented by x, is fixed by the action of H iff xH/x = J.

By the fixed point principle, the number of cosets of J fixed by H is congruent to m mod p. This is nonzero, so some coset xJ is fixed by H. This means H and J are conjugate through x. Yet the selection of J was arbitrary, so every sylow subgroup is a conjugate of H, and all sylow subgroups are conjugate.

All sylow subgroups are isomorphic. Conjugation by the appropriate element implements the isomorphism.

A sylow subgroup is normal iff it is the only sylow subgroup in G.

Cascading Sylow Groups

Let a group G have a p sylow subgroup S, with k conjugates. Let C be a normal subgroup of G such that p does not divide |C|. Let H be the quotient group G/C. Map S into H, giving a subgroup T. Nothing of order p can live in C, so S and C are disjoint. This means S maps injectively into H. In other words, S and T are isomorphic.

Suppose S₁ and S₂ map to the same sylow group T in H. This means the C cosets of S₁ contain S₂. If you know that S₁ is normal in the group generated by C and S₁, then S₁ is the only sylow subgroup within this group. S₂ cannot be different from S₁. The sylow subgroups of G and H correspond one for one. This is the case when C is the center of G. Show that S₁ is normal in C*S₁, and you're all set. You can then repeat this process, pulling out the center of H, and so on, as long as the centers are not divisible by p.

With the sylow group unchanged, and the number of sylow groups unchanged, we are really pulling subgroups out of the normalizer at each step. This is clear when C is the center, for it certainly normalizes S.