Dtyjlui

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 $2$.