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EDIT: the $p$-stable definition I give below is incorrect. I have included the correct definition as an answer to this question.


I am trying to understand the definition of a p-stable group. The first part of the definition is

A faithful representation of a finite group $G$ on a vector space over a field of characteristic $p\not= 2$ is called $p$-stable if no $p$-element of $G$ has a quadratic minimal polynomial.

What does it mean for a group element to have a minimal polynomial?

Additionally, any intuition on a meaningful interpretation of this definition would be much appreciated. What is special about elements which have quadratic minimal polynomials? Why would we want to get rid of them? What's wrong with $p=2$?

After this,

  1. If $G$ has no nontrivial $p$-subgroups, $G$ is $p$-stable if every faithful characteristic $p$ representation is $p$-stable.

  2. If $1<O_p(G)$ and $1=O_{p'}(G)$ then $G$ is $p$-stable if for all normal nontrivial $p$-subgroups $P$, for every $p$-element $x$ such that $[[x,P],x]=1$, the image $\overline{x}$ in $G/C_G(P)$ is contained in a normal $p$-subgroup.

  3. If $1<O_p(G)$ and $1<O_{p'}(G)$, then $G$ is $p$-stable if $G/O_{p'}(G)$ is $p$-stable.

What, mainly, is the connection between the $p$-stable representation definition and $\#2$? Are these somehow the same, but in a different light?

(I see that $[[x,P],x]$ are elements of the form $p^{-1}x^{-1}pxx^{-1}x^{-1}p^{-1}xpx=(x^{-1})^p(p^{-1})^xpx$, so if that is equal to $1$ then $px=p^xx^p$. So there's sort of a "double twist" happening, which must be important in some way; but I don't see immediately any connection to minimal polynomials.)

Sorry if these are basic questions on advanced material. I am sure the answer to this part is, to some extent, because this is a technical definition which is made to prove things with, but even the broadest intuition on this would help.

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I don't have any intuition on the definition. But presumably minimum polynomial of a group element $g \in G$ with respect to a representation $\phi$ refers to the minimum polynomial of $\phi(g)$. –  JSchlather Feb 6 '13 at 20:31
    
I think, from the phrase "satisfying a technical condition introduced by Gorenstein and Walter ... in order to extend Thompson's uniqueness results in the odd order theorem" this definition came backwards - it was defined to be useful based on an existing argument that worked in other cases. I can't be sure of that, but that feels likely. –  Thomas Andrews Feb 6 '13 at 20:54
    
@ThomasAndrews Maybe somebody who knows how these things were used in the classification theorems can help. If it is backwards like that, it would at least be useful to know what they were trying to do with it. –  Samuel Handwich Feb 6 '13 at 20:58
    
That wikipedia page is also somewhat suspect - it has no links to external definitions. (I thought $p$-element would be an element of order $p$, but later definitions would be meaninngless if that was true .) –  Thomas Andrews Feb 6 '13 at 21:05
    
@ThomasAndrews I think a $p$-element is supposed to be an element of $p$-power order. –  Samuel Handwich Feb 6 '13 at 21:08

4 Answers 4

up vote 1 down vote accepted
What is special about elements which have quadratic minimal polynomials?

As the action of a $p$-element $x\in G$ on a (finite dimensional) vector space $V$ over a field of characteristic $p$ is nilpotent, defining $V_0 = V, V_{i+1} := [V_i, x]$ you get $V_n = 0$ for some $n\ge 0$. As $x$ acts trivially if $n\le 1$, the minimal nontrivial case is $n=2$, i.e., $x$ acts quadratically.

 Why would we want to get rid of them?

An action of $G$ on $V$ is $p$-stable if for all $a \in G$ holds $$[V, a, a] = 1 \implies a\mathrm{C}_G(V) \in \mathrm{O}_p(G/\mathrm{C}_G(V)).$$ The purpose of this condition is to exclude sections of $G/\mathrm{C}_G(V)$ that are isomorphic to $\mathrm{SL}_2(p)$ (and act "naturally" like $\mathrm{SL}_2(p)$, see for example the remark after Theorem 9.1.4 in [KS]). If you have a group $G$ with $\mathrm{C}_G(\mathrm{O}_p(G)) \le \mathrm{O}_p(G)$ and the action of $G$ on the chief factors of $G$ in $\mathrm{O}_p(G)$ is $p$-stable then Glauberman's ZJ-Theorem states that $Z(J(S))$ is normal $G$ for every $p$-Sylow subgroup $S$ of $G$ (see beginning of section 9.4 in [KS]).

What's wrong with p=2?

Every element of order $2$ acting non-trivially on a vector space over a field of characteristic $2$ acts quadratically (according to the remark after 9.4.5 in [KS] you can replace $p$-stability by excluding sections of $G$ isomorphic to $\mathrm{SL}_2(2) = S_3$ directly to get meaningful results for $p=2$).


[KS] Kurzweil, Stellmacher: The Theory of Finite Groups

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As @ThomasAndrews suggested, the definition given on Wikipedia was not correct. I have located the definition in an article by Glauberman:

  1. Let $p$ be an odd prime and $G$ be a finite group with $O_p(G)\not= 1$. Then $G$ is $p$-stable if it satisfies the following condition: Let $P$ be an arbitrary $p$-subgroup of $G$ such that $O_{p'}(G)P$ is a normal subgroup of $G$. Suppose that $x\in N_G(P)$ and $\overline{x}$ is the coset of $C_G(P)$ containing $x$. If $[P,x,x]=1$, then $\overline{x}\in O_n(N_G(P)/C_G(P))$.

  2. Define $\mathcal{M}_p(G)$ as the set of all $p$-subgroups of $G$ maximal with respect to the property that $O_p(M)\not= 1$.

  3. Let $G$ be a finite group and $p$ an odd prime. Then $G$ is called $p$-stable if every element of $\mathcal{M}_p(G)$ is $p$-stable.

(I am not sure what little $n$ is in (1). I am guessing just any $n\in \mathbb{N}$?)

So I guess that $[P,x,x]$ is a commutator condition that no critical $p$-subgroups are allowed to have when we want $G$ to be $p$-solvable. (Evidently, the case $p\not= 2$ is degenerate with respect to the $[P,x,x]$ condition, which is why it is not included in this.) As a side note, excluding $SL(2,p)$ from a group implies that it is $p$-stable - I suspect because the $[P,x,x]$ relation gives rise to that sort of subgroup.

I am still not sure what to think intuitively about what makes these groups "stable" with respect to $p$. Perhaps I will find and read the papers in the bibliography.

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If you can get the book The Theory of Finite Groups by Kurzweil and Stellmacher, take a look at chapter 9 "Quadratic Action". –  j.p. Feb 6 '13 at 21:50

That Wikipedia page is a mess.

If there are no $p$-subgroups of $G$, then there are no $p$-elements of $G$, and therefore, by the definition of $p$-stable representation, all faithful representations are $p$-stable, since a representation being $p$-stable is defined only in terms of how it affects $p$-elements, and there are none.

There are lots of indications that this Wikipedia page is not "mature." Specifically, there are no links to external articles defining the terms, and there are cases of mangled grammar.

I would not trust that page. Probably best to go to the primary source.

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I have found a paper in my library which includes the definition, which has some differences. Should I rewrite this question or start a new one? –  Samuel Handwich Feb 6 '13 at 21:17
    
It depends on your intent. I think you have the answer to your "technical" question: What does it mean by the minimal polynomial? The bigger question of "why this definition?" is still open, and it does seem like a separate question. –  Thomas Andrews Feb 6 '13 at 21:19

I don't study these things so I can't give you any intuition, but here is the definition of the minimal polynomial:

Given a representation $\phi\colon G \to \mathrm{GL}_n(k)$ and an element $g \in G$ the matrix $A = \phi(g)$ is square so given any polynomial

$$f(x) = c_nx^n + c_{n-1}x^{n-1} + \cdots c_1x + c_0$$

it is perfectly well defined to set

$$f(A) = c_nA^n + c_{n-1}A^{n - 1} + \cdots + c_1A + c_0I$$

where $I$ is the $n \times n$ identity matrix. The minimal polynomial of $g$ with respect to $\phi$ is then the monic polynomial $f$ of minimal degree such that $f(A) = 0$ (the $n \times n$ matrix whose entries are all zero).

Also: we can define the ideal $\mathrm{ann}(A)$ of all polynomials $p$ such that $p(A) = 0$. The minimal polynomial $f$ is the monic polynomial that generates this ideal. So if $p(A) = 0$ then you can write $p = fh$ for some other polynomial $h$.

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