# What is the intersection of all Sylow $p$-subgroup's normalizer?

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.

• – Jack Schmidt Feb 17 '14 at 23:21
• Nice argument,$[R,P]\leq O_p(G)$ – mesel Feb 18 '14 at 13:38
• 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. – Jack Schmidt Feb 18 '14 at 14:48
• I have observed that $P\cap R$=$O_p(G)$ for any sylow p subgroup. – mesel Feb 18 '14 at 15:54
• 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 $P$. – Jack Schmidt Feb 18 '14 at 16:11

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
• 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 $N_G(P^g)$). – Mikko Korhonen Feb 21 '14 at 19:21
• (Sorry the examples were from another question of mesel. I'm redoing the census now.) – Jack Schmidt Feb 21 '14 at 19:44
• 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 $R$? – mesel Feb 21 '14 at 19:56
• The new census is still running, but it suggests R=Z and Q≤Z are not the standard. – Jack Schmidt Feb 21 '14 at 21:05
• @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. – Jack Schmidt Feb 21 '14 at 21:14

$R_p(G)$ 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.

• :Thanks again Jack. – mesel Mar 25 '14 at 15:10

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

• 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. – Jack Schmidt Feb 22 '14 at 6:26
• :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 $O_p(G)$. – mesel Feb 22 '14 at 9:10
• @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. – Jack Schmidt Feb 22 '14 at 15:42
• 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). – Jack Schmidt Feb 22 '14 at 21:51
• 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 ? – mesel Feb 22 '14 at 22:13