What is a minimal polynomial of a group element, and why would we care if it was quadratic? 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,


*

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


*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.


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


*

*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 $\overline{x}\in O_n(N_G(P)/C_G(P))$.


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


*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.
