I asked this question on Stackexchange, but I got no answer within 24 hour, so I ask it here.
Iwasawa's criterion can be stated in the two following forms :
1° Let $G$ be a perfect group acting faithfully and primitively on a set $X$ with a least two elements. Assume that there exists an element $x$ of $X$ such that the stabilizer $G_{x}$ of $x$ has an abelian subgroup $A$ which is normal in $G_{x}$ and such that the conjugates of $A$ in $G$ generate $G$. Then $G$ is simple.
2° Let $G$ be a perfect group acting primitively on a set $X$ with a least two elements. Assume that there exists an element $x$ of $X$ such that the stabilizer $G_{x}$ of $x$ has an abelian subgroup $A$ which is normal in $G_{x}$ and such that the conjugates of $A$ in $G$ generate $G$. Then the quotient of $G$ by the kernel of the action of $G$ on $X$ is simple.
By use of Iwasawa's criterion, one can prove the following statement :
3° if $n$ is a natural number $\geq 2$, if $F$ is a field, if we don't have simultaneously $n = 2$ and $\vert F \vert \leq 3$, then $PSL(n, F)$ is simple.
Statement 3° can also be deduced from the following theorem :
4° if $n$ is a natural number $\geq 2$, if $F$ is a field, if we don't have simultaneously $n = 2$ and $\vert F \vert \leq 3$, if $L$ is a normal subgroup of $SL(n, F)$ not contained in the center of $SL(n, F)$, then $L = SL(n, F)$.
If I'm not wrong, 4° cannot be obained as a straightforward consequence of 3°, thus 4° is "stronger" than 3°.
Robinson, A Course in the Theory of Groups, 2d edition, p. 74-78, gives a proof of 4° which doesn't rely on Iwasawa's criterion (and not even on group actions) but which involves many calculations difficult to memorize. In fact, analyzing the manner in which 3° is deduced from Iwasawa's criterion in Rotman, 4th edition, 1999, p. 279, one sees that 4° can be easily deduced from the following strengthening of Iwasawa's criterion :
5° Let $G$ be a perfect group acting primitively on a set $X$. Assume that there exists an element $x$ of $X$ such that the stabilizer $G_{x}$ of $x$ has an abelian subgroup $A$ which is normal in $G_{x}$ and such that the conjugates of $A$ in $G$ generate $G$. If $L$ is a normal subgroup of $G$ not contained in the kernel of the action of $G$ on $X$, then $L = G$.
This strengthening of Iwasawa's criterion can be proved by a straightforward extension of the proof of Iwasawa's criterion given in Rotman, p. 263-264. (I sketch the proof below.)
My question is : do you know a reference to the literature for the strengthening of Iwasawa's criterion (i.e. for 5°) ? Thanks in advance.
Here is a sketch of the proof of statement 5°.
Lemma 1. Let $G$ be a group acting primitively on a set $X$, let $H$ be a normal subgroup of $G$ not contained in the kernel of this action. Then the action of $H$ on $X$ is transitive.
Proof. Easy extension of the proof of Rotman, 4th edition, 1999, theor. 9.17, (i), p. 258. Our Lemma 1 can also be deduced from this theorem by considering the associated faithful action.
Lemma 2. Let $G$ be a group, let $H$ and $S$ be subgroups of $G$ such that $G = HS$. If $K$ is a normal subgroup of $S$ whose conjugates in $G$ generate $G$, then $G$ is generated by $H$ and $K$. If in addition $H$ is normal in $G$ and $K$ abelian, then $H$ contains the commutator subgroup of $G$.
Proof. Implicitly in Rotman, proof of theor. 9.27, p. 263-264.
Proof of 5°. By Lemma 1, the action of $L$ on $X$ is transitive. By the generalized Fattini argument (Rotman, exerc. 4.9 (i), p. 81), $G = L G_{x}$, where $x$ is as in the hypothesis of 5°. By hypothesis of 5°, $G_{x}$ contains an abelian subgroup $A$, normal in $G_{x}$ and such that the conjugates of $A$ in $G$ generate $G$. Thus, by Lemma 2, $L$ contains the commutator subgroup of $G$. But $G$ is assumed to be perfect, thus the commutator subgroup of $G$ is the whole $G$, thus $L = G$.
