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.



Finite groups factorized into two simple alternating groups
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](https://www.dropbox.com/s/2jglyu6nuhrnws5/alternatingproduct.pdf), 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.


