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Let $G$ be a finite solvable group with an abelian Sylow $p$ subgroup $S$.

A classic theorem of Burnside says that $S$ has a normal complement if and only if the Centralizer of $S$ in $G$ is equal to the Normalizer of $S$ in $G$. Since the centralizer is contained in the normalizer we can just look at the index. So for any group $G$ let ${\rm icn}_p(G)$ be the index of the centralizer of a Sylow $p$ subgroup in its normalizer.

First just consider $p=2$, but much of this, if not all, could be independent of $p$. I would like to know how ${\rm icn}$ behaves w.r.t. subgroups.

If $G$ is solvable with abelian Sylow $2$ subgroup $S$ is it true for every subgroup $H \subset G$ that ${\rm icn}(H) \le {\rm icn}(G)$? Moreover is it true that ${\rm icn}(H)$ is a divisor of ${\rm icn}(G)$?

These seem like they should be elementary, but I don't see how to approach.

A quick computation shows that if solvability and abelian are not required, then the second statement about divisors is false, but the first inequality still seems to hold. An example of this is the symmetric group $S_4$. Here the Sylow 2 subgroup is not abelian, but for each subgroup of $K$ of $S_4$, $icn(H)$ is still not greater than 4. In fact $icn(S_4) = 4$, and $A_4 \subset S_4$ gives $icn(A_4) = 3$

Is this true for any finite group $G$ and subgroup $H \subset G$ that ${\rm icn}(H) \le{\rm icn}(G)$?

Thanks for your help

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    $\begingroup$ What is your example where the divisor statement fails? $\endgroup$ Aug 23, 2021 at 20:25
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    $\begingroup$ (The reason I ask is that every $G$-automorphism of a subgroup of an abelian Sylow $p$-subgroup extends to a $G$-automorphism of the whole Sylow $p$-subgroup, so I think the second statement holds for all finite groups.) $\endgroup$ Aug 23, 2021 at 22:57
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    $\begingroup$ In your fourth paragraph you seem to be using $H$ for two different things: an abelian Sylow $2$-subgroup of $G$, and an arbitrary subgroup of $G$. $\endgroup$ Aug 24, 2021 at 3:17
  • $\begingroup$ I edited the question to remove both abelian and solvable. Sorry about that. And in the last paragraph I clarified the role of $H$. And I see it needs another edit. H should never mean a Sylow subgroup. $\endgroup$
    – vginhi
    Aug 24, 2021 at 22:18

1 Answer 1

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What you are looking for is the idea of control of fusion. Let $S$ be a Sylow $p$-subgroup of $G$, and let $H$ be a subgroup of $G$ containing $S$. We say that $H$ controls fusion in $G$ with respect to $S$ if, whenever $A$ and $B$ are two subsets of $S$ that are conjugate in $G$ via some element $g$, then there exists $h\in N_G(S)$ such that $A^h=B$ and, furthermore, conjugation by $g$ and $h$ induce the same function on $A$.

Burnside proved that if $S$ is abelian then $N_G(S)$ controls fusion in $S$ with respect to $G$. In general, if $N_G(S)$ controls fusion then $N_H(T)/C_H(T)$ is isomorphic to a subgroup of $N_G(S)/C_G(S)$, whenever $H$ is a subgroup and $T$ is a Sylow $p$-subgroup of $H$. So your second claim holds whenever $N_G(S)$ controls fusion. This is really a statement about $p$-groups: a $p$-group $S$ where $N_G(S)$ always controls fusion when $S$ is a Sylow in $G$ are called resistant. Almost all finite $p$-groups are resistant.

Counterexamples to your general claim, that $\mathrm{icn}(H)\leq \mathrm{icn}(G)$, abound for simple groups, particularly for the prime $2$, when $\mathrm{icn}(G)$ can equal $1$. In this case, Frobenius's normal $p$-complement theorem states that $G$ has a normal $p$-complement if and only if $\mathrm{icn}(H)$ is a $p$-group for all subgroups $H$. If we choose a group without a normal $2$-complement, but with $N_G(S)=SC_G(S)$, then we should be able to find a counterexample.

For a concrete one, let $G=\mathrm{SL}_2(9)$ and $H=\mathrm{SL}_2(3)$ for $p=2$. Then $\mathrm{icn}(G)$ has order $8$, because $N_G(S)=S$ and the group is quaternion. The Sylow $2$-subgroup of $H$ is also quaternion but of order $8$. However, it is normal in $H$, so $\mathrm{icn}(H)=|H|/2=12$.

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  • $\begingroup$ icn is the index of the centralizer of the Sylow 2 subgroup in the normalizer of the Sylow 2 subgroup. So for $A4$, we have S = D_8, with centralizer C_2, and normalizer $D_8$. So we get a quotient of 4. In the case of $A_4$, the Sylow subgroup is abelian, and its Centralizer is itself $C_2 \times C_2$, while the normalizer is $A_4$, so the index is 12/4 =3. This is still less than 4. $\endgroup$
    – vginhi
    Aug 25, 2021 at 23:18
  • $\begingroup$ @vginhi You are right that I have messed up with the numbers. I'll just change the counterexample. $\endgroup$ Aug 26, 2021 at 8:37
  • $\begingroup$ Basically the idea is to choose $G$ with a self-normalizing Sylow $2$-subgroup $S$. Then your quantity is $|S|/|Z(S)|$. We know that $S$ possesses subgroups $T$ whose normalizer is not a $2$-group, so must have an element of order at least $3$ in the normalizer. If the subgroup is of index $2$ in $S$ and $|Z(S)|=|Z(T)|$ then your quantity will increase. I just made the mistake of choosing $T$ abelian rather than non-abelian. Doing so fixes the problem that $|Z(T)|$ jumps. $\endgroup$ Aug 26, 2021 at 8:45
  • $\begingroup$ Thanks for this. It does seem that self normalizing sylow 2 groups are not so common. My quick search seems to indicate that self normalizing sylow 2 groups cannot be solvable groups. That might say that the statement holds for solvable groups. Is that correct. And does the divisible statement still hold if S is not abelian, but G is solvable? It works for the dihedral groups, for example. $\endgroup$
    – vginhi
    Aug 27, 2021 at 19:58
  • $\begingroup$ Almost all soluble groups have self-normalizing Sylow $2$-subgroups. For example, $S_3$ and $S_4$. And $G=\mathrm{GL}_2(3)$ with $H=\mathrm{SL}_2(3)$ is a soluble counterexample, with the same Sylow $2$-structure as $\mathrm{SL}_2(9)$. The statement isn't true in general, even for soluble groups. $\endgroup$ Aug 27, 2021 at 20:07

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