# Possible Converse to Iwasawa Lemma

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.

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.

• 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? 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)

Then:

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, $$A_n\triangleright\text{ker}(\phi)=\{e\}$$