I have two isomorphic subgroups $U,V\leq G$ and an element $a \in U \cap V$. Is it true that $U$ and $V$ are conjugate under $G$ if and only if they are conjugate under $N_G(\langle a \rangle)$? I vaguely remember hearing such a statement but at the moment I can neither think of a proof nor find a counterexample. Thanks in advance for any help.

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    $\begingroup$ One question I think I never really answered too well was: Suppose U,V are G-conjugate p-subgroups of G. Set d(U,V) = 1 if there is some non-identity p-subgroup Q such that U,V are $N_G(Q)$-conjugate. It is not necessarily true that d(U,V) = 1 (I think even assuming U,V are abelian). How large can d(U,V) be? (We define a graph whose vertices are the G-conjugates of U, and the edges are all U-V such that d(U,V) = 1 and U≠V. The distance is then the length of the shortest path from U to V.) $\endgroup$ Feb 29, 2012 at 18:10
  • $\begingroup$ @Jack: This is related to the question of the minimal size of a conjugation family (see work of Alperin, Goldschmidt etc). The conjugation family defined by Goldschmidt circa 1970 was essentially minimal. I am unsure, but the name of S. Dolan may be relevant to your question. $\endgroup$ Feb 29, 2012 at 18:59
  • $\begingroup$ @Geoff: I found in a few examples that though the conjugation family had 4 or 5 members, the distance on the graph was actually always 1 (you had to use different Q for different U,V, but there was always a Q that worked). I believe it was GL or related, and the N_G(Q) were the maximal subgroups containing the Sylow (I forget if these are maximal parabolics or minimal parabolics, but the ones that are maximal in terms of subgroup inclusion). So I've been confused how large d(U,V) could be (as I only know a few groups with large conjugation families). $\endgroup$ Feb 29, 2012 at 23:07
  • $\begingroup$ @Jack: The paper of S.Dolan is reviewed as MR0430067. I think in Lie type characteristic $p$ groups, the minimal size of a conjugation family is the Lie rank, which is also the number of minimal parabolics (counting the whole group as a minimal parabolic in rank 1). If you use all the fusion from a maximal parabolic, then you can efect quite a lot. $\endgroup$ Feb 29, 2012 at 23:26
  • $\begingroup$ @geoff: thanks. And you've reminded me of the problem with my examples: we're only supposed to use $N_G(Q)$ to conjugate inside Q. I'll check the paper, but I suspect that means minimal parabolics are needed, and d will be close to the size of the conjugation family. $\endgroup$ Feb 29, 2012 at 23:34

2 Answers 2


No, this is false as stated. For example, let $G = {\rm GL}(3,2).$ Let $a$ be an involution of $G.$ Let $U = C_{G}(a),$ which is dihedral of order $8.$ Let $V$ be another Sylow $2$-sbgroup of $G$ in which $a$ is not central. (This is possible, because all involutions in $G$ are conjugate, but not all involutions of $U$ are in $Z(U)).$ Now $U$ and $V$ are certainly conjugate within $G,$ as both are Sylow $2$-subgroups of $G.$ They are certainly not conjugate via an element of $N_{G}(\langle a \rangle) = C_{G}(a) = U.$

  • $\begingroup$ Thank you very much. I had a feeling that the statement should be false but couldn't find a counterexample. Do you happen to know if the result changes if we assume $U$ to be Abelian? $\endgroup$ Feb 29, 2012 at 14:45
  • $\begingroup$ If $U$ is Abelian, and is Sylow $p$-subgroup of $G,$ then $U$ and $V$ are clearly already conjugate within $C_G(a).$ Without some sort of Sylow condition, I think that the answer is no in general, even if $U$ is Abelian. $\endgroup$ Feb 29, 2012 at 14:55

(To answer a question in the comments with an easy example)

This is not even true when U is abelian. Conjugation of p-subgroups is controlled by local subgroups, but you cannot be quite so picky about which local subgroups you use (in particular, $N_G(\langle a\rangle)$ may be quite small).

Let $G=\operatorname{AGL}(1,8) \cong 7 \ltimes (2\times2\times2)$ be the group of invertible affine transformations $x\mapsto ax+b$ for $a,x,b$ in GF(8) and $a≠0$. There are many Klein four-subgroups of G, but their normalizers are all the (normal) Sylow 2-subgroup P of order 8, $\{ x \mapsto x+b : b \in \operatorname{GF}(8)\}$. For instance the subgroup U generated by $x\mapsto x+1$ and $x\mapsto x+\zeta$ for $\zeta$ a primitive 7th root of unity is conjugate in the whole group G to the subgroup V generated by $x \mapsto x+1$ and $x\mapsto x+\zeta^2$. However, the normalizer of $x\mapsto x+1$ is the Sylow 2-subgroup P, and the two subgroups are obviously not conjugate inside an abelian group P.

In particular, the local subgroups are the Sylow 2-subgroup P as well as the entire group G. The normalizer of every proper non-identity subgroup of P is P itself (since $G/P$ acts faithfully and irreducibly on P and is cyclic of prime order), so most local subgroups are P. However, the normalizer of P itself is all of G and so it is this local subgroup that really controls the fusion.

Similar examples work for any prime p and elementary abelian Sylow p-subgroup P with proper subgroups U and $U^g$ that have P as the normalizer of their intersection (or any non-identity subgroup of their intersection).

For abelian Sylow p-subgroups, the normalizer of the Sylow always controls the fusion (Burnside's theorem), but the normalizer of proper p-subgroups may be much too small to control anything.


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