# Finite groups factorized into two simple alternating groups

My research is somehow related to the following question :

Describe and classify all finite groups $G$ such that $G=HK$ with $H \cap K=1$, where $H \cong A_m$ and $K \cong A_n$ for some integers $m, n$ greater than 4. ($A_n$ denotes alternating group of degree $n$).

In a paper of O. Kegel and H. Luneberg, (Uber die kleine reidemeister bedingungen, Arch. Math. (Basel) 14 (1963)), all groups $G$ with the factorization $G=AB$ with $A \cong B \cong A_5$ are classified. I need to know about other progresses have been done on the above question. How much is known about this problem in general? Any references and hints would be appreciated. Thanks in advance.

• @StefanKohl I just make it more clear what I mean. Some updates on results relate with the problem is also added. – Farrokh Shirjian Jun 5 '14 at 16:57
• @Stefan: $G = HK$ is a stronger condition than $G$ being generated by $H$ and $K$. The OP is asking for all Zappa-Szep products of alternating groups (en.wikipedia.org/wiki/Zappa%E2%80%93Sz%C3%A9p_product). – Qiaochu Yuan Jun 5 '14 at 17:35
• @Qiaochu: Now, after the edit, the question is clear. Therefore I removed my first comment. – Stefan Kohl Jun 5 '14 at 22:18

For $m=5$ and $n$ arbitrary, all factorizations are classified here:

W. R. Scott, Products of $A_5$ and a finite simple group, J. Algebra 37 (1975), 165--171.

For $m=6$ and $m=7$ (again with $n$ arbitrary) all factorizations are classified by a series of papers by Darafsheh and various coauthors. The relevant titles are:

M. R. Darafsheh and G. R. Rezaeezadeh, Factorization of groups involving symmetric and alternating

M. R. Darafsheh and G. R. Rezaeezadeh and G. L. Walls Groups which are the product of $S_6$ and a simple group

M. R. Darafsheh, Product of the symmetric group with the alternating group on seven letters

M. R. Darafsheh, Factorization of simple groups involving the alternating group

For $m=8$ the problem is solved when $G$ is simple:

Deqin Chen, Products of simple group involving the alternating $A_8$

Update

1. In fact, there is a paper by Darafsheh that addresses this problem for arbitrary $m$ and $n$:

Darafsheh, M. R. Finite groups which factor as product of an alternating group and a symmetric group. Comm. Algebra 32 (2004), no. 2, 637–647.

In this paper Darafsheh proves that the group $G$ is either $H\times K$ or is an alternating group, although he does not specify of what degree. (You could work this out from the write-up below though.)

1. I wrote up the answer I gave earlier, which combined with ideas from Geoff Robinson's answer, yields the following result:

Theorem: Suppose that $G$ is a finite group containing two proper subgroups $H$ and $K$ such that $G=H.K$. Suppose, moreover that $H\cong{\rm Alt}(m)$ and $K\cong{\rm Alt}(n)$ with $m,n\geq 5$. Then one of the following holds: (1) $m,n\leq 80$; (2) $G={\rm Alt}(n+1)$; (3) $G={\rm Alt}(m)\times{\rm Alt}(n)$.

Note that case (1) accounts for only a finite number of possible groups $G$. The theorem asserts, therefore, that, barring a finite number of cases, $G$ is isomorphic to the groups in cases (2) and (3).

Case (3) is, of course, entirely explicit and encompasses an infinite family of examples. Consider, on the other hand case (2): In this case $K$ is the stabilizer of a point in the natural action of $G={\rm Alt}(\ell)$ on $\ell$ points. Thus $H$ must be a transitive subgroup of $G$ and any factorization $G=K.H$ arises from the action of $H$ on the set of cosets of one of its subgroups. The problem of classifying all such factorizations is equivalent to determining all (conjugacy classes of) subgroups of $G={\rm Alt}(\ell)$.

Note that if one imposes the supposition that $H\cap K=\{1\}$ in the statement of the theorem (as the OP originally asked), then one can say more: The extra supposition allows one to strengthen the statement of (2) to assert that $n+1=m!$ thereby making this case entirely explicit. To see why this extra assertion follows we use the remarks of the previous paragraph and observe that, given the extra supposition, the corresponding transitive action of $H$ must be simply transitive.

• Sounds great! I just need some more time to think...Thanks so much to you and Geoff. – Farrokh Shirjian Jun 7 '14 at 6:55
• I think you meant $n+1 = m!$, as you said in your comment on my post. – Geoff Robinson Jun 8 '14 at 0:06
• @Geoff, yes you are right. I'll edit this post in due course, as I believe I can now prove the same proposition without the requirement that $G$ be simple - one needs to also allow for the possibility that $G=K\times H$ in this case. I'll write it up when I have chance. – Nick Gill Jun 9 '14 at 19:45
• Yes, also the general argument I wrote up later in my post shows in particular that if the simple factors have trivial intersection in the simple homomorphic image, then that simple group admits no proper extensions with the same property. – Geoff Robinson Jun 9 '14 at 22:52
• Note that the condition $n+1=m!$ in the statement of the proposition requires that $H\cap K=\{1\}$. – Nick Gill Jun 10 '14 at 21:46

I think that the assumption that $A_{m} \cap A_{n} = 1$ may get in the way. You may need to drop it for inductive reasons. For example, you can take a group $G$ such that $G = A_{m}A_{n}$ for subgroups $A_{m},A_{n}$, with $G$ of minimal order subject to that. Such a group must either be simple (assuming $m,n >4$), or else one of the factors must be normal. For if $N \lhd G$ with $N$ neither trivial nor $G,$ then either $G/N$ would have the same property, contrary to the choice of $G,$ or else $N$ would contain one of the alternating factors. With the current state of knowledge, I would have thought that the simple case could be eliminated, or at least fully described if, it can happen at all. So, even if you start off with a group $G = A_{m}A_{n}$ where you insist on $A_{m} \cap A_{n} = 1,$ if that group is not simple, it would be natural to look at homomorphic images. A maximal normal subgroup $M$ of $G$ would either contain one of the factors, or else $G/M$ would be a simple (non-Abelian) group which is a product of alternating groups, but possibly with non-trivial intersection. Note that if $M$ contains $A_{n},$ then we would have $G/M \cong A_{m}.$ But then by order considerations, we would have $M = A_{n},$ in which case $G = A_{n} \times A_{m}$ is forced by consideration of ${\rm Aut}(A_{n}).$ Hence the problem seems to reduce to understanding simple groups expressible in the form $A_{n}A_{m}$ where some intersection is allowed.

Later remarks: This seems to be general strategy for classifying finite groups expressible in the form $G = ST$ with $S,T$ non-Abelian simple. If we take a maximal normal subgroup $M$ (in the case that $G$ itself is not simple), we find either that ${\bar G} = G/M$ is a simple group with ${\bar G} = {\bar S}{\bar T},$ and ${\bar S} \cong S$ and ${\bar T} \cong T,$ or else (using the-now proved via the Classification of Finite Simple Groups - Schreier conjecture) that $G \cong S \times T.$ Furthermore, in the former case, we note that $|{\bar S} \cap {\bar T}| = |M||S \cap T|,$ so that, in particular, $|M|$ divides $|{\bar S} \cap {\bar T}|$. Hence, even if there is a finite simple group ${\bar G}$ with such a factorization, there are a finite number of possible extensions $G$ which admit a similar factorization, and the largest order of a group by which ${\bar G}$ can be extended is bounded by ${\rm gcd}(|S|,|T|).$

• surely order considerations are going to greatly restrict when a simple group can be the product of two alternating groups. For instance any group of Lie type has a Sylow subgroup that has order at least a cube root the order of the group. That alone is probably enough to (pretty much) rule out groups of Lie type.... So only the alternating groups will be tricky... – Nick Gill Jun 5 '14 at 18:47
• ... and for the alternating groups, I would think that the main example will be when $n+1=m!$ and one considers a regular action of $A_m$ on itself (so $G=A_{n+1}$).... This aside, I imagine there will only be sporadic examples – Nick Gill Jun 5 '14 at 20:40
• @Nick Gill : I agree, I think it's just a matter of checking. – Geoff Robinson Jun 5 '14 at 23:24
• I think I can prove that the examples I mention are the only examples when $G$ is simple (up to a finite number of exceptions). I'll try and write something down in due course... – Nick Gill Jun 6 '14 at 2:37
• @FarrokhShirjian : Yes,I think so, although though I think that Nick Gills answer shows that with finitely many exceptions, we either have $G = H \times K$ or $H \cap K =1.$ – Geoff Robinson Jun 10 '14 at 18:55