Sylow Subgroups and Conjugation

Suppose that $$G$$ is a finite group whose order is divisible by a prime $$p$$. Let $$S$$ be the set of Sylow $$p$$-subgroups of $$G$$; let $$H$$ be an element of $$S$$. $$H$$ acts on $$S$$ by conjugation. The fact that is the key to the proof of the Third Sylow Theorem is that there is only one $$H$$-orbit of order $$1$$. I was wondering if we could say more about this action.

• I first thought that maybe there are only two $$H$$-orbits. But this turned out to be false in general. (I saw that it was false in the dihedral group $$D_{5}$$.)
• Next I checked whether all the orbits other than $$\{H\}$$ has order $$H$$. This is false in general, too. (This is false in $$A_{5}$$.)

After fiddling with some examples, I found that

all the $$H$$-orbits except $$\{H\}$$ seem to have the same order.

But I can neither prove or disprove this claim. So here are my questions:

1. Is the above claim correct? If so, why?
2. Is there any other things we can say about the $$H$$-orbits?

Any help is appreciated. Thanks in advance!

Let $$\mathcal{S}=\{S \leq G: S \in Syl_p(G)\}$$ be the set of Sylow $$p$$-subgroups of $$G$$. Fix a $$S \in \mathcal{S}$$ and let $$S$$ act on $$\mathcal{S}$$ by conjugation. Then the length of an orbit of a $$T \in \mathcal{S}$$ is clearly $$|S:N_S(T)|$$, the index of the normalizer of $$T$$ relative to $$S$$. The following holds true.

Proposition Let $$S,T$$ be Sylow $$p$$-subgroups of $$G$$, then $$N_S(T)=S \cap T=N_T(S)$$.

Proof Let's prove the first equality, since by symmetry the other holds to. Clearly if $$x \in S \cap T$$, then $$x \in N_S(T)$$. So assume $$x \in N_S(T)$$, in particular $$x \in S$$. Then $$\langle x \rangle T$$ is a subgroup. And, it is a $$p$$-subgroup, since $$|\langle x \rangle T|=\frac{|\langle x \rangle| \cdot |T|}{|\langle x \rangle \cap T|}$$ and note that $$x$$ is a $$p$$-element. But $$T \subseteq \langle x \rangle T$$, but $$T$$ is a maximal $$p$$-subgroup since it is Sylow. Hence $$T = \langle x \rangle T$$, that is, $$x \in T$$.

So the size of each of the orbits is $$|S:S \cap T|$$ (which equals $$|T:S \cap T|$$). Your question boils down to what can be said about the intersections of the different Sylow $$p$$-subgroups. These do not have to be equal as the example of $$S_3 \times S_3$$ and its Sylow $$2$$-subgroups demonstrate (see @Verret). Finally observe, since the oribit size of $$S$$ itself is $$1$$ as you remarked, the number Sylow $$p$$-subgroups $$n_p(G) \equiv 1$$ mod $$|S:S \cap T|$$, where $$|S \cap T|$$ is chosen to be as large as possible among the $$T$$'s not equal to $$S$$. This generalizes the regular formula $$n_p(G) \equiv 1$$ mod $$p$$.

By computer calculation, the smallest counterexample is $$S_3^2$$ which has $$9$$ Sylow $$2$$-subgroups, and the action of one of them by conjugation has one fixed point, one orbit of length $$4$$, and two orbits of size $$2$$.

• Thank you for your answer! I accepted Nicky's answer but yours was equally helpful. I have one question; what software/program did you use? (I have never used computers for calculations involving groups and I am curious if there is any suggestion.) – Ken Jan 10 '19 at 1:28
• I used "magma", which is not free, but they have a free online calculator at magma.maths.usyd.edu.au/calc which is more than good enough for easy calculations like this. A free alternative is "GAP" gap-system.org – verret Jan 10 '19 at 3:04
• Just to get you started, this is a program that runs through the groups of order at most 10 and prints the order of their sylow subgroups: for n in [1..10] do for i in [1..NumberOfSmallGroups(n)] do G:=SmallGroup(n,i); for p in PrimeDivisors(Order(G)) do P:=SylowSubgroup(G,p); [n,i,p]; #P; end for; end for; end for; – verret Jan 10 '19 at 3:07