I conjecture that:
Every Finite Homomorphic image of an infinite (with arbitrary cardinality) product of finite solvable groups is solvable -- or at least Not a simple (non-abelian) group.


I can see this conjecture in some cases.
but the general case seems very complicated.


Question: Has this problem been investigated ?
Thank you. 


Edit-- Moreover, I also conjecture that Every Finite Homomorphic image of an INFINITE product G = \prod G\_i of finite simple (non-abelian) groups {G\_i}, is a product of finite simple groups each of which is isomorphic to some G\_i.


Y Cor has kindly outlined the proof of my 2 conjectures (in comments below) via the Nikolov-Segal Theorem (Annals of Math., 2007). Moreover, Yilong Yang has 
just found another proof (of my 2nd conjecture) by using his covering properties of finite groups (introduced in his publication in 2016). 



Finite Homomorphic images of infinite products of finite solvable groups
That's correct: every finite quotient is solvable. Indeed let $G$ be your product of finite solvable groups. Let $p:G\to F$ be a (possibly non-continuous) surjective homomorphism to a finite group $F$. Lift $F$ to a finite subset $\tilde{F}$, and let $H$ be the closed subgroup generated by $\tilde{F}$. Then by the Nikolov-Segal theorem, the restriction of $p$ to $H$ is continuous. But as a closed subgroup of $G$, $H$ is a projective limit of finite solvable groups. So all its quotients by open subgroups are finite solvable groups.


More generally, this shows that 



> 
> any abstract finite quotient of a profinite group $G$ is isomorphic to some $H/K$ where $H$ is a closed subgroup of $G$ and $K$ an open normal subgroup of $H$.
> 
> 
> 


(In the particular case when $G$ is pro-$p$ for some prime $p$, this appeals to a much older theorem of Serre instead of Nikolov-Segal.)




---


(Edit, to answer a question in a comment.) The previous actually shows a slightly stronger statement: 



> 
> (1) if $G$ is a profinite group and $F$ is a finite quotient of $G$, then there exists a t.f.g. closed subgroup $M$ of $G$ such that for every t.f.g. closed subgroup $H$ of $G$ containing $M$, $F$ is isomorphic to the quotient of $H$ by some closed normal subgroup.
> 
> 
> 


[t.f.g. means topologically finitely generated, meaning having a dense finitely generated subgroup]


Now I claim that 



> 
> (2) if $G$ is a product $\prod S\_i$ of finite simple groups (possibly abelian), then every t.f.g. closed subgroup of $G$ is contained in a t.f.g. closed subgroup that is a product of finite simple groups.
> 
> 
> 


If we combine (1) and (2), we deduce that 



> 
> in an arbitrary product of finite simple groups $\prod\_{i\in I}S\_i$, every abstract finite quotient is isomorphic to some finite product $\prod\_{j\in J}S\_j$
> 
> 
> 


(although it's not always given by the obvious projection!).


Actually while (1) follows from the Nikolov-Segal theorem (every finite index subgroup is open, in a t.f.g. profinite group), the particular case of the Nikolov-Segal theorem for products of finite simple group was proved by Saxl and Wilson, namely: if $(S\_n)$ is a sequence of finite simple groups and $\lim\_{n\to\infty}|S\_n|=\infty$, then every finite index subgroup in $\prod S\_n$ is open. (J. Saxl and J. S. Wilson, A note on powers in simple groups, Math. Proc. Cambridge Phil.
Soc. 122 (1997), 91-94.)


Checking (2) is quite elementary, modulo the fact that finite simple groups have a bounded generating rank, namely every finite simple group is generated by 2 elements, which relies on classification. Gather together all isomorphic simple groups in the product, to write $G=\prod\_S G\_S$, where each $G\_S\simeq S^{I\_S}$, and $S$ ranges over non-isomorphic finite simple groups. Let $L$ be a t.f.g. closed subgroup of $G$; so $L$ has a dense subgroup generated by $d$ elements for some $d$. Let $L\_S$ be its projection in $S^{I\_S}$; it is generated by $d$ elements. Let $M\_S$ be the subgroup of $S^{I\_S}$ generated by $L\_S$ and the diagonal; it is generated by $\le d+2$ elements (because any finite simple group is generated by 2 elements). Also, every finitely generated subgroup of $S^{I\_S}$ containing the diagonal is isomorphic to $S^k$ for some $k$. So if we choose a family of $d+2$ elements in $\prod M\_S$ whose $S$-component generates $M\_S$, we obtain a generating subset for a dense subgroup of $\prod M\_S$, and $M\_S$ is a product of finite simple groups. This proves (2). 


