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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!

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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 $p$.

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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 $2$.

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  • $\begingroup$ 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.) $\endgroup$ – Ken Jan 10 at 1:28
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    $\begingroup$ 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 $\endgroup$ – verret Jan 10 at 3:04
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    $\begingroup$ 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; $\endgroup$ – verret Jan 10 at 3:07

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