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Every cyclic p-group is a Sylow p-subgroup of a finite group whose distinct Sylow p-subgroups intersect trivially in pairs (and there is at least one pair).

For instance, let q be a prime congruent to 1 mod p and then the group $\operatorname{AGL}\left(1, q\right)$ is such a group. This is the normalizer of a Sylow q-subgroup in the symmetric group on q points and can be described as the group of affine transformations of a line over the field of q elements:

$$\operatorname{AGL}\left(1, q\right) = \left\{ x \mapsto ax + b \mid a, b \in \mathbb{Z}/q\mathbb{Z}, ~~a \ne 0 \right\}.$$

Its Sylow p-subgroups are the cyclic subgroups Pb generated by xzx + b where z is a primitive pth root of unity in Z/qZ. They intersect trivially, since Pb leaves b/(1−z) alone, but moves every other point.

I can handle a few other cases (quaternion, elementary abelian), but I haven't found a (correct) general method. I was a little surprised semi-direct products with faithful (even irreducible) modules was insufficient.

Which p-groups are Sylow p-subgroups of finite groups whose distinct Sylow p-subgroups intersect trivially in pairs, and there is at least one pair?

In other words, though every pair of distinct Sylow p-subgroups of P intersects trivially, it does so vacuously with no pairs. For each p-group P, I want a finite group G with more than one Sylow p-subgroup $P$ where $P\cap P^g$ in $\{ 1, P \}$ for all $g\in G$.

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For $p=2$ take a look at jstor.org/pss/1970491 –  j.p. Sep 23 '11 at 7:41
@jug: Thanks! I summarized this in my CW answer. –  Jack Schmidt Sep 23 '11 at 14:22
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2 Answers

up vote 4 down vote accepted

Your question can be rephrased as

which $p$-groups appear as non-normal T.I. Sylow $p$-subgroups?

Aside: T.I. stands for "trivial intersection". A trivial intersection set is one that intersects each of its conjugates fully or trivially. These have been actively studied, because groups that have T.I. sets exhibit some interesting representation theoretic behaviour (see e.g. Chapter 7 of Isaacs). I am sure that Jack knows all this, I am writing it for the benefit of other readers. [/Aside]

I believe that Lemma 1.1, and Propositions 1.2 and 1.3 in this paper answer your question. Essentially, either $P$ is cyclic, or generalised quaternion, or the question is reduced to T.I. Sylow $p$-subgroups of simple groups by virtue of Proposition 1.2 (b), (d), and (h). Namely, (h) and (b) say that usually $P\leq U$, and then (d) reduces everything to simple groups (you quotient out the $p'$-core, so the isomorphism class of the $p$-Sylow doesn't change). At that point, one just has to go through the classification, which is done in Proposition 1.3.

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Thanks! I'm still tracking down some of the references. –  Jack Schmidt Sep 23 '11 at 14:22
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Here are my notes, since I like fairly explicit answers.

In case anyone wants to see the other examples more briefly:

Elementary abelian p-groups of order pn are found TI in PSL(2,pn). Such subgroups are uniquely identified by which subspace they stabilize.

Quaternion 2-groups are found TI in the their semi-direct products with their (unique) faithful irreducible module over a prime field of odd order. The key point is that the unique element of order 2 acts fixed-point-freely on the module, and so is not contained in the intersection of any two distinct Sylow 2-subgroups.

A weaker question has a positive answer:

For every p-group P is there a finite group G such that for some g in G, PPg = 1?

Yes. For any p-group P, take G to be the semi-direct product of P with a large enough faithful module V over a finite field of characteristic not p. Then consider the union of CV(x) as x varies over P. Since V is faithful, each centralizer is a proper subspace, and if V has large enough dimension, it cannot be written as the union of |P| proper subspaces. If v is some element of V outside that union of centralizers, then PPv = 1.

For p = 2, the classification is due to Suzuki (1964) who discovered his infinite family of finite simple groups in this same line of investigation. Not only did he classify the possibly P, but also the possible G. Let N be the largest odd-order normal subgroup of G.

  • P is cyclic and G = PN
  • P is quaternion of order 8, and either G = PN, or G / N ≅ SL(2,3)
  • P is generalized quaternion and G = PN
  • P is elementary abelian of order 2n and PSL(2,2n) ≤ G / N ≤ PΓL(2,2n)
  • P is the Sylow 2-subgroup of PSU(3,2n) and PSU(3,2n) ≤ G / N ≤ PΓU(3,2n) — P is sort of a GF(2n) version of the quaternion group of order 8.
  • P is the Sylow 2-subgroup of Sz(2n) and G / N = Sz(2n)

In other words, I missed two infinite families (and one fusion system). The structure of N is restricted, but not I think not classified. In some sense, one should read this list as "there exists an N such that...".

The classification for odd p appears to be a post-CFSG result, and is closely related to strongly embedded subgroups (similar to the concept Arturo mentions below: malnormal). The strongly embedded list is on page 383 - 384 of number 3 of the GLS writeup of the CFSG. The TI list is in Blau-Michler (1990), but is based on older lists I have not yet tracked down.

Other than cyclic and elementary abelian, there are a few more infinitely families, and a few sporadic exceptions.


  • Suzuki, Michio. "Finite groups of even order in which Sylow 2-groups are independent." Ann. of Math. (2) 80 (1964) 58–77. MR162841 DOI:10.2307/1970491

  • Blau, H. I.; Michler, G. O. "Modular representation theory of finite groups with T.I. Sylow p-subgroups." Trans. Amer. Math. Soc. 319 (1990), no. 2, 417–468. MR957081 DOI:10.2307/2001249

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I think there is a name for the property that $H\cap H^g$ is either $H$ or trivial; but I can't seem to remember what it is. (I thought it might be "malnormal", but that's a stronger condition, since it says that $H^g\cap H=1$ for all $g\notin H$, so you also require $H$ to be self-normalizing). –  Arturo Magidin Sep 23 '11 at 3:53
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