Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question is partly influenced by the question:

Are two subgroups that contain a common element conjugate iff they are conjugate under the normalizer?

If we have an arbitrary finite group $G$ and radical $p$-subgroups $A,B\leq G$ (so $A=O_{p}(N_{G}(A),p)$, $B=O_{p}(N_{G}(B),p)$ and the normalizers of $A$ and $B$ in $G$ are thus parabolic subgroups) such that $A\leq B\leq N_{G}(B)\leq N_{G}(A)\leq G$ (where $p\in\pi(G)$), is it the case that there is a unique conjugate of $A$ contained in $B$?

Clearly $A$ is normal in $B$, but I cannot seem to prove that there cannot be an elemnet $g\in G\setminus N_{G}(A)$ such that $A^{g}\ne A$ and $A^{g}\leq B$. Equally I have not found an example where uniqueness does not hold. Any hints would be greatly appreciated.

Edit: Initially the question did not state that $A$ and $B$ were radical subgroups, as a greater understanding of the more general situation as deemed to be desirable.

share|cite|improve this question
Why the tags combinatorics and definition? – joriki Aug 10 '12 at 10:30
@joriki. The definitions tag was a mistake - I meant to use the finite group tag, but somehow got the definition one by mistake. The combinatorics tag is because I am looking at the different ways that conjugates of subgroups respect subgroup inclusions. However, feel free to remove this tag if you deem it to be irrelevant. – David Ward Aug 10 '12 at 10:33
You are trying to prove that $A$ is weakly closed in $B$. Look up this term (or "weak closure") in any group theory book. Often the case $B$ a $p$-Sylow subgroup is of interest. – j.p. Aug 10 '12 at 12:59
You could also reread Geoff's answer to the question you linked to... – j.p. Aug 10 '12 at 13:34
@jug. Thanks for the advice. The example given in Geoff's answer does indeed give a counterexample to my question as stated above. My area of interest is with radical subgroups as implicitly defined in Geoff's joint paper with Reinhard Knorr of 1989 (MR0989918), but it is nevertheless a useful example to bear in mind. – David Ward Aug 10 '12 at 14:56
up vote 6 down vote accepted

It is true for groups of Lie type in defining characteristic

A unipotent subgroup of a connected, reductive linear algebraic group is called radical if it is the unipotent radical of its normalizer. One version of the Borel–Tits theorem is that the radical subgroups are the unipotent radicals of the parabolic subgroups. Hence the radical subgroups of a fixed Borel subgroup are indexed by subsets of the positive roots, and so we have the rather convenient $A \leq B \implies N_G(B) \leq N_G(A)$ whenever $A,B$ are radical subgroups of a connected, reductive linear algebraic group. Alperin's fusion theorem (for unipotent subgroups of a connected, reductive linear algebraic group) shows that whenever $\langle A,A^g \rangle$ is unipotent, there are radical subgroups $B_i$ and elements $g_i \in N_G(B_i)$ such that $A^{g_1 \cdots g_{i-1}} \leq B_i$ and $g = g_1 \cdots g_n$. In particular, if $A$ itself is radical and $\langle A,A^g \rangle$ is unipotent, then $A=A^g$ since all the $g_i \in N_G(B_i) \leq N_G(A)$. In other words, every radical subgroup of a connected, reductive linear algebraic group is weakly closed in every maximal unipotent subgroup containing it.

It is false in general

However, when $G$ has a characteristic different from $p$ (or has no characteristic), then things loosen up considerably. We no longer have $A \leq B \implies N_G(B) \leq N_G(A)$ for radical subgroups, and this allows us to focus attention on the “wrong” $B$, while another $B$ takes care of the conjugation. $\newcommand{\Sym}{\operatorname{Sym}}\newcommand{\PSL}{\operatorname{PSL}}$

A few infinite families of counterexamples for $p=2$ include $\Sym(n)$ for at least half of $n\geq 7$, and $\operatorname{PSL}(3,q)$ for $|q| \geq 5$ odd. I assume counterexamples are quite common in general, but there appear to be a lot of “small exceptions” to my assumption, so beware.

For $G=\Sym(7)$, let $A=\langle (1,2) \rangle$, which has order 2 and normalizer $A \times \Sym(\{3,4,5,6,7\})$ and let $B=\langle (1,2), (3,4), (5,6), (3,5)(4,6) \rangle$ be a self-normalizing Sylow 2-subgroup. Clearly $A$ has many $G$-conjugates in $B$, including $\langle (3,4) \rangle$ and $\langle (5,6) \rangle$. This does not contradict Alperin's fusion theorem, since the conjugation that takes $A$ to $\langle (5,6) \rangle$ does not have to rely only on the subgroup $B$. A similar construction works for most $n$: if $2,3 \equiv n \mod 4$ then more or less the exact construction works. I am a little unclear on how easily $\Sym(9)$ works while $\Sym(8)$ does not work at all.

For $G=\PSL(3,q)$, ($|q| \geq 5$ an odd prime power, $\PSL(3,-|q|) = \operatorname{PSU}(3,|q|)$ being the unitary group), there are two cases for the proof, but the choice of $A$ and $B$ is the “same.” Take $B$ to be a Sylow 2-subgroup and $A=Z(B)$ to be its center. For $q=1 \mod 4$, $B$ is a “wreathed” Sylow 2-subgroup, and $A=Z(B)$ is the “diagonal” subgroup of the base of $B$. $A$ is $G$-conjugate to all the direct factors of the base, but obviously is normal in $N_G(B)$. Again this does not contradict Alperin's fusion theorem, since $A$ is contained in the base of $B$, a radical subgroup whose normalizer does not normalize $A$. For $3 \equiv q \mod 4$, $B$ is quasi-dihedral and $A=Z(B)$ has order 2. $A$ is contained in a radical Klein 4-subgroup with full fusion, but any larger radical subgroup has smaller normalizer. I'm not entirely certain I understand why $q=3$ is an exception.

share|cite|improve this answer
Thanks for such a thorough answer. The condition $A\leq B\Rightarrow N_{G}(B)\leq N_{G}(A)$ is the key. Thanks also for giving an explanation as to why it works for groups of Lie Type in defining characteristic as this is an area I am not as familiar with. – David Ward Aug 11 '12 at 20:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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