Thanks for any help or comments.

Suppose that $G$ is a finite group. A Carter subgroup of $G$ is a nilpotent self-normalizing subgroup of $G$. Carter and Vdovin have shown that solvable groups have Carter subgroups, and that in addition, in every group with Carter subgroups, the Carter subgroups are conjugate -- see

Carter, R. W. (1961), Nilpotent selfnormalizing subgroups of soluble groups, Mathematische Zeitschrift, 75 (2): 136–139.

Vdovin, E. P. (2006), On the conjugacy problem for Carter subgroups. (Russian.), Sibirsk. Mat. Zh., 47 (4): 725–730. Translation in Siberian Math. J. 47 (2006), no. 4, 597–600

Vdovin, E. P. (2007), Carter subgroups in finite almost simple groups. (Russian.), Algebra i Logika, 46 (2): 157–216.

My question is about the structure of groups whose Carter subgroups are their Sylow $2$-subgroups? I mean, I am interested in any theorem which guides me towards some classification of this type of groups.

  • 1
    $\begingroup$ You are asking to classify finite groups with a self-normalizing Sylow $2$-subgroup. The so-called $C(G,T)$-Theorem of M. Aschbacher may be relevant here. $\endgroup$ Nov 18, 2016 at 15:40
  • 1
    $\begingroup$ To expand: a first step might be to consider finite groups in which Sylow $2$-subgroup is maximal, where the Theorem of Aschbacher mentioned above shoudl certainly be helpful. $\endgroup$ Nov 18, 2016 at 15:43
  • $\begingroup$ @Geoff, thaks. I am not familar with Aschbacher theorem. At least do you think is it possible to characterize groups are minimal with respect to this property? I mean groups contain self-normalizing 2-sylow subgroup but every subgroup does not contain such subgroup. $\endgroup$
    – maryam
    Nov 18, 2016 at 17:55
  • 1
    $\begingroup$ I think that that these minimal groups in particular have a maximal Sylow $2$-subgroup, and I do think that groups with a maximal Sylow $2$-subgroups may be attackable as I suggested above. $\endgroup$ Nov 18, 2016 at 18:02
  • 1
    $\begingroup$ A structure theorem for finite nonsolvable groups with a maximal Sylow $2$-subgroup was proved by Bernd Baumann, J. Alg. 38 #1 (1976). $\endgroup$ Mar 11, 2017 at 16:39

2 Answers 2


Classifying finite groups with self-normalizing Sylow $2$-subgroup seems rather hopeless as $10608361$ of the $10625619$ groups of order less than $768$ have this property. --

A GAP function to count the groups of order $n$ with this property is as follows:

NrOfGroupsWithSelfNormalizingSylow2Subgroup := function ( n )

  if   n = 1 then return 1;
  elif SmallestRootInt(n) = 2 then
    return NrSmallGroups(n);
  elif n mod 2 = 1 then
    return 0;
    return Number(AllGroups(n),
                  G -> Size(SylowSubgroup(G,2))
                     = Size(Normalizer(G,SylowSubgroup(G,2)))); 
  • 1
    $\begingroup$ Thanks. I think this number is rather amazing, since it is possible to exclude all p-groups of order less than 768. $\endgroup$
    – maryam
    Nov 18, 2016 at 18:35
  • 1
    $\begingroup$ Almost all of those groups are the groups of order 512, though, which is a potentially misleading result. I don't think it makes sense to apply the question at hand to p-groups. It's trivial in those cases. $\endgroup$ Nov 18, 2016 at 18:53

Here's some stronger evidence in favor of Stefan's claim that such groups are too numerous and varied to expect a classification.

glist := AllSmallGroups(Size,[2..511],IsPGroup,false,G->Size(G) mod 2 = 0, true,
Sum(List(Filtered([2..511],n->(n mod 2 = 0) and 
    (not IsPrimePowerInt(n))),n->NrSmallGroups(n)));

This shows that of the 33510 non-p-groups with even order less than 512, 25673 of them have self-normalizing Sylow 2-subgroups. Moreover, given $n\leq 7$, every 2-group of order $2^n$ appears as the Sylow 2-subgroup in some such example.

ForAll([1..7],n->NrSmallGroups(2^n) = Number(syls,P->P[1]=2^n));

So not only does it look like we can expect a large proportion of groups to have this property, but that we cannot even reasonably expect to constrain the structure of the Sylow 2-subgroup.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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