After looking at how Iwasawa's Lemma proves the simplicity of $PSL_n(q)$ and $PSp_{2n}(q)$ groups, I wondered how powerful this lemma is. As such, I thought about this possible converse.

For reference, Iwasawa's Lemma:

For group $G$,

  1. $G$ has a primitive group action $\phi:G\rightarrow\text{Sym}(X)$ on set $X$ (transitive, and $\text{Stab}(x)$ is maximal subgroup of $G$)
  2. $G=G'$ (commutator subgroup)
  3. There is a soluble normal subgroup $A$ of $\text{Stab}(x)$ such that $G=\langle gAg^{-1}|g\in G\rangle$.

Then $G/\text{ker}(\phi)$ is simple.

A possible converse can be postulated:

For every simple group $G$, $G$ can be expressed as $H/N$, where

  1. $H$ has a primitive group action $\phi:H\rightarrow\text{Sym}(X)$ on set $X$ (transitive, and $\text{Stab}(x)$ is maximal subgroup of $H$)
  2. $H=H'$ (commutator subgroup)
  3. There is a soluble normal subgroup $A$ of $\text{Stab}(x)$ such that $H=\langle hAh^{-1}|h\in H\rangle$.
  4. $N=\text{ker}(\phi)$

The overarching question is: Which simple groups can possibly satisfy the converse of Iwasawa's Lemma?

You may assume Classification of Finite Simple Groups.

Here's a possible attempt:

As we know the alternating groups $A_n (n\geq5)$ and cyclic groups of prime order $C_p$ are simple, we can try to look at them first. However, the natural action $A_n$ acting on $\{1...n\}$ does not work because the stabilisers are isomorphic to $A_{n-1}$ with no nontrivial normal subgroups for $n\geq 6$.

On examination, it follows that $G$ must act transitively on $X$ too. However, for $G=C_p$, this means that $|X|=p$ otherwise the action is trivial. I'm not sure how to further analyse this case, but it doesn't seem possible for abelian simple groups.

It was observed that $A_6$ can act on left cosets of $S_4$ (with inclusion map $\sigma\mapsto\sigma$ if $\sigma$ is even and $\sigma(5,6)$ otherwise) in $A_6$ via left multiplication. $A_6$ be expressed as $A_6/\{e\}$ with a maximal subgroup isomorphic to $S_4$ (standard twisted $S_4$). $S_4$ trivially has a soluble normal subgroup $A_4$, containing the 3-cycle $(1,2,3)$, whose conjugates generate $A_6$. This doesn't seem to generalise easily to the other altermating groups.


2 Answers 2


Let $G$ be a non-abelian finite simple group. Then $G = G'$ and there exists a maximal subgroup $M < G$ with a normal solvable subgroup $N \neq 1$. In other words, the converse to Iwasawa's theorem holds.

See the answer and comments to this question on MO: Maximal subgroups of simple groups with normal $2$-subgroups.

  • $\begingroup$ For completion, the $L_p(3)$ (or $PSL_p(3)$?) groups seem to have a weaker property than a normal 2-subgroup, do you have any reference for that? $\endgroup$
    – Element118
    Dec 25, 2020 at 5:04

Here is a partial answer for alternating groups $A_n,n\geq5,n\neq6$.

$A_n$ can be expressed as $A_n/\{e\}$. We take the subgroup $S$ of permutations that preserve $\{1...n-3\}$, i.e. $\{\sigma(1)...\sigma(n-3)\}=\{1...n-3\}$. Note:

  1. $A_n=A_n'$, since $A_n'\trianglelefteq A_n$ and $A_n'$ is nontrivial.
  2. $S$ is maximal in $A_n$ (conjectured with GAP)


  1. Let $A_n$ act on left cosets $gS$ in $A_n$ by left multiplication. Then $\text{Stab}(eS)=S$ is maximal and the action $\phi:A_n\rightarrow\text{Sym}(gS)$ is transitive.
  2. $S\triangleright\langle(n-2,n-1,n)\rangle$, this subgroup is soluble and conjugates of it generate $A_n$ (since 3-cycles generate $A_n$). This requires $n\geq5$.
  3. Clearly, $A_n\triangleright\text{ker}(\phi)=\{e\}$

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