I'm working on understanding when and why $\text{PSL}(2,q)$ is simple and I've encountered Iwasawa's lemma, although I'm struggling to understand part of the proof. I'm referring to Wilson's book on the finite simple groups, where he states the lemma as follows:
If $G$ is a finite perfect group, acting faithfully and primitively on a set $\Omega$, such that the point stabilizer $H$ has a normal abelian subgroup $A$ whose conjugates generate $G$ then $G$ is simple.
My first question is, why the wording "the point stabilizer $H$?" Is there some condition we are assuming about $G$ which allows for a unique stabilizer subgroup, up to isomorphism? I don't think this is the case, as other wordings of Iwasawa's lemma will say something along the lines of "such that some point stabilizer H..."
My main confusion is in the proof, and it's in someways the same confusion as the above paragraph. The proof is by contradiction, so he first assumes $G$ is not simple, hence there exists some non-trivial normal subgroup $K$. Using the fact that the action is faithful, and that $K$ is non-trivial, we deduce that there is a stabilizer subgroup $H$ which does not contain $K$. But then, Wilson begins speaking of $A$, a normal abelian subgroup of $H$. I'm confused as to why we know this specific $H$ is the one mentioned in the hypothesis of the Lemma, and how we can know this $A$ exists.
Thanks in advance.
A point-stabilizer subgroup depends on a choice of point. However $\mathrm{Stab}(gx)=g\mathrm{Stab}(x)g^{-1}$ implies all points in an orbit have conjugate stabilizers. Thus, we can speak of "the" point-stabilizer up to conjugacy. If any point-stabilizer $H$ has a normal abelian subgroup $A$ whose $G$-conjugates generate $G$, then any conjugate subgroup $gHg^{-1}$ also has a normal abelian subgroup $gAg^{-1}$ whose $G$-conjugates generate $G$. The exercise is assuming an arbitrary representative of a conjugacy class of subgroups is chosen.