11
$\begingroup$

Intersection of all Sylow $p$-subgroups is generally denoted by $O_p(G)$ and it is one of the well studied topics in group theory as there are many theorems related to this.

Let $R$ be intersection of all Sylow $p$-subgroup's normalizer in $G$. It is easy to observe that $R$ is a characteristic subgroup of $G$ containing $O_p(G)$. I wonder the properties of $R$ and its relation with $O_p(G)$.

If anyone can find or observe something about $R$, I would be thankful.

$\endgroup$
  • 2
    $\begingroup$ See math.stackexchange.com/questions/18467/… $\endgroup$ – Jack Schmidt Feb 17 '14 at 23:21
  • $\begingroup$ Nice argument,$[R,P]\leq O_p(G)$ $\endgroup$ – mesel Feb 18 '14 at 13:38
  • $\begingroup$ I haven't had any luck (or much time) getting anything more specific though. The last time I investigated, I didn't find any papers on it. $R$ is the kernel of the permutation/conjugation action of $G$ on its Sylow $p$-subgroups, so is definitely important. $\endgroup$ – Jack Schmidt Feb 18 '14 at 14:48
  • $\begingroup$ I have observed that $P\cap R$=$O_p(G)$ for any sylow p subgroup. $\endgroup$ – mesel Feb 18 '14 at 15:54
  • $\begingroup$ That's very good. That means $O_p(G)$ is the normal Sylow $p$-subgroup of $R$, so $R=Q \ltimes O_p(G)$ for some $p'$-subgroup $Q$. Now we (by which I probably mean you :-) just need to figure out what $Q$ looks like, either in terms of $G$ or in terms of its action on $O_p(G)$ or on $P$. $\endgroup$ – Jack Schmidt Feb 18 '14 at 16:11
5
+300
$\begingroup$

For reference, here is what we've got so far:

Let $G$ be a finite group with Sylow $p$-subgroup $P$ and set $R= \bigcap N_G(P^g) = \bigcap N_G(P)^g$ to be the intersection of the Sylow $p$-normalizers.

$R$ is an important subgroup, it is the normal core of $N_G(P)$ and the kernel of the permutation action of $G$ on its Sylow $p$-subgroups. In particular, if one wanted to organize groups by how they acted on their Sylow $p$-subgroups, we'd need two invariants: (1) a transitive permutation group whose point stabilizer is the normalizer of a Sylow $p$-subgroup, and (2) $R$.

Consider $[P,R]$. Since $R$ normalizes $P$, we get $[R,P] \leq P$. However $R$ is itself a normal (characteristic) subgroup of $G$, so $[R,P] \leq R$ as well. In other words $[R,P] \leq R \cap P$.

Consider $R \cap P$, a $p$-subgroup normalizing each Sylow subgroup $P^g$. Since $(R \cap P) P^g$ is a subgroup, it is a $p$-subgroup, and so is actually equal to $P^g$, since $P^g$ is maximal amongst $p$-subgroups. Hence $R \cap P \leq P^g$ for all $g$. Taking the intersection we get $R \cap P \leq O_p(G)$. Since $O_p(G)$ is a $p$-subgroup of $R$ contained in $P$, we also get $O_p(G) \leq R \cap P$. Hence $R \cap P = O_p(G)$.

For any $G$-normal subgroup $X$, $X \cap P$ is a Sylow $p$-subgroup of $X$. Hence $O_p(G)$ is a normal $p$-subgroup of $R$. By Schur-Zassenhaus $R=Q \ltimes O_p(G)$ for some $p'$-subgroup $Q$. Since $[Q,P] \leq R\cap P = O_p(G)$, we get that $Q$ centralizes $P/O_p(G)$, but $Q$ need not centralize $O_p(G)$, lest it centralize all of $P$.

Indeed, I think one of the first things to decide is how much different $R$ is from $Z=\bigcap C_G(P^g) = \bigcap C_G(P)^g$. When $O_p(G)=1$, we get $R=Z$, so that $Q \leq Z$.


A survey of small groups (in progress) reveals a variety of structures of $R/Z$:

Amongst the isomorphism classes of groups $G$ with $|G|\leq 1000$ and the conjugation action of $G$ on its Sylow 3-subgroups isomorphic to $A_4$'s action, the quotients $R/Z$ occur with the following frequencies:

  • $R=Z$, 1705 times
  • $[R:Z] = 2$, 199 times
  • $[R:Z] = 3$, 115 times
  • $R/Z \cong C_4$, 5 times
  • $R/Z \cong C_2 \times C_2$, 13 times
  • $R/Z \cong S_3$, 49 times
  • $R/Z \cong C_6$, 3 times
  • $R/Z \cong C_8$, 1 times
  • $R/Z \cong D_8$, 1 times
  • $R/Z \cong Q_8$, 1 times
  • $R/Z \cong C_3 \times C_3$, 151 times
  • $R/Z \cong C_3 \times S_3$, 2 times
  • $R/Z \cong \operatorname{Dih}(C_3 \times C_3)$, 17 times
  • $R/Z \cong C_3 \ltimes C_9$, 9 times
  • $R/Z \cong C_3 \ltimes (C_3 \times C_3)$, 26 times
  • $R/Z \cong C_3 \times C_3 \times C_3$, 21 times
  • $R/Z \cong \operatorname{Dih}(C_3 \times C_3 \times C_3)$, 1 times
$\endgroup$
  • $\begingroup$ You could also see $R \cap P = O_p(G)$ from $N_G(P^g) \cap P = P^g \cap P$ (since $P^g$ is the unique Sylow subgroup of $N_G(P^g)$). $\endgroup$ – Mikko Korhonen Feb 21 '14 at 19:21
  • $\begingroup$ (Sorry the examples were from another question of mesel. I'm redoing the census now.) $\endgroup$ – Jack Schmidt Feb 21 '14 at 19:44
  • 1
    $\begingroup$ I wonder something,Even if $O_p(G)$ is intersection of all sylow-p subgroup, it is known that when $G$ is solvable, intersection of three suitable sylow-p subgroup is $O_P(G)$.Can we say smilar argument for normalizer of sylow-psubgroup and $R$? $\endgroup$ – mesel Feb 21 '14 at 19:56
  • $\begingroup$ The new census is still running, but it suggests R=Z and Q≤Z are not the standard. $\endgroup$ – Jack Schmidt Feb 21 '14 at 21:05
  • $\begingroup$ @mesel: your intersection question can be answered inside $G/R$, and I think is called the size of the basis or stabilizer chain of the permutation group (an intersection of conjugates of $N_G(P)$ is an intersection of stabilizers, and so we are looking for a short list of points whose pointwise stabilizer is the trivial subgroup). So far no counterexamples. $\endgroup$ – Jack Schmidt Feb 21 '14 at 21:14
2
$\begingroup$

$R_p(G)$ is $p$-solvable, but for every odd prime $p$ there is a finite group $G$ such that $R_p(G)$ is not solvable.

We consider solvability properties of $R_p(G) = \bigcap\{ N_G(P^g) : g \in G \}$ where $P$ is a Sylow $p$-subgroup of $G$. By the previous answer, $R_p(G) = Q \ltimes O_p(G)$ is $p$-closed, so definitely $p$-solvable ($p$-length 1, even).

If $p=2$, then clearly $R_2(G)$ is solvable by Feit–Thompson's odd order theorem. In cases where $|G|\leq 1000$, $R_p(G)$ is always solvable. However, in general this need not be true, since we can take $R_p(G) = G$ by taking any $G$ with a normal Sylow $p$-subgroup. Any such group is $p$-solvable, but need not be solvable. For example $G=A_5 \times C_7$ and $p=7$ works.

A slightly less trivial example (the smallest pefect example in fact) is $G=A_5 \times \operatorname{GL}(3,2)$ with $p=5$ or $p=7$ and $R_p(G)$ is the coprime direct factor. The next smallest perfect example is $G=\operatorname{SL}(2,5) \ltimes \operatorname{GF}(11)^2$ with $p=11$, and then $O_p(G)=P=Z:=\bigcap\{ C_G(P^g) : g \in G\}$ is a Sylow $p$-subgroup, and $R_p(G)/O_p(G)=\operatorname{SL}(2,5)$ is $11$-solvable (being of order coprime to 11) but not solvable.

$\endgroup$
  • $\begingroup$ :Thanks again Jack. $\endgroup$ – mesel Mar 25 '14 at 15:10
2
$\begingroup$

On the embedding of $R$ in $G$:

$R$ is similar to $O_p(G)$, and it is a fairly convenient fact that $O_p(G) = P \cap P^g$ for two Sylow $p$-subgroups of $G$ for many $G$ (for instance $G$ with abelian Sylow $p$-subgroups by Brodkey (1963), or $G$ $p$-solvable for $p>2$ by Itô (1958) and Robinson (1984)), and $O_p(G) = P \cap P^g \cap P^h$ for all finite $G$ by Mazurov-Zenkov (1995). In other words, the intersection of all Sylow $p$-subgroups is also the intersection of a few well chosen Sylow $p$-subgroups.

$R$ is the intersection of all Sylow $p$-normalizers, so it is a reasonable question whether $R$ is the intersection of just a few Sylow $p$-normalizers.

Itô's specific bound of 2 is not enough for $R$: Let $G_1= \operatorname{GL}(2,2) \times \operatorname{GL}(2,2)$, $V= \operatorname{GF}(2)^4$ its natural module, and $G=G_1 \ltimes V$ be the associated affine group. Then the natural action of $G$ (with $V$ a regular normal subgroup) is also the action of $G$ by conjugation on its Sylow 3-subgroups. By Itô (1958), $O_p(G)$ is the intersection of two of its Sylow $3$-subgroups, but a direct calculation shows $R$ is not the intersection of two Sylow $3$-normalizers (but is the intersection of 3).

There are 6 other examples with very similar behavior ($p$ odd, $G$ is $p$-solvable, $R$ is the intersection of three but not two Sylow $p$-normalizers; in each case $p=3$ and $G$ is actually solvable).

Brodkey's bound of $2$ is also not enough for $R$: let $G=A_5 \times D_{10}$ acting on its Sylow $2$-subgroups (not its natural action). Then $O_p(G)$ s equal to the intersection of (any) two Sylow $2$-subgroups, but again $R$ requires three Sylow $2$-normalizers. There are four other examples of $G$ with less than 30 Sylow $2$-subgroups, all of which are abelian, yet whose $R$ is not the intersection of any two Sylow $2$-normalizers; in each case $R$ is the intersection of three Sylow $2$-normalizers.

Bibliography

$\endgroup$
  • $\begingroup$ This leaves open the possibility of a slightly larger bound. It would be nice to know if 4 is enough or if there is an unbounded sequence. $\endgroup$ – Jack Schmidt Feb 22 '14 at 6:26
  • $\begingroup$ :As far as I know,To say that three sylow-$p$ subgroups are enough to find $O_p(G)$ you must require $G$ is solvable.(there is no general bound proved).And Brodkey proved that if a sylow-p subgroup is abelian then two of them are enough to find $O_p(G)$. $\endgroup$ – mesel Feb 22 '14 at 9:10
  • $\begingroup$ @mesel: fixed. I include $O_p(G)$ citations and for Itô's and Brodkey's versions, examples where the analogues for $R$ don't hold. $\endgroup$ – Jack Schmidt Feb 22 '14 at 15:42
  • 1
    $\begingroup$ Hrm, not sure how I missed it earlier but the product action $S_4 \wr S_2$ is the action on the Sylow $3$-subgroups by conjugation, and it requires 4 Sylow 3-normalizers. It is of course solvable. examples with abelian Sylows seem common ($\operatorname{P\Gamma L}(2, 8)$ for instance). $\endgroup$ – Jack Schmidt Feb 22 '14 at 21:51
  • 1
    $\begingroup$ The last thing which I wonder:$F(G)$ is Product of $O_P(G)$ for all-p dividing $G$.it is known that $F(G)$ is largest nilpotent normal subgroup of $G$.Let denote $R_p$ instead of $R$ for a fixed prime $p$.And set R(G) as product of all $R_p$.if $R(G)\neq F(G)$ then $R(G)$ can not be nilpotent.Can you observe anything about it? Should it be solvable ? $\endgroup$ – mesel Feb 22 '14 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.