# Finite quotients of an infinite product of finite groups

Let $G$ be a finite group. Consider the direct product $\Gamma=\prod_{i=1}^{\infty}G$ of (countably) infinitely many copies of $G$. For every finite set of numbers $\{i_1,\ldots,i_n\}$ we have the natural projection $\phi_{i_1,\ldots,i_n}:\Gamma\rightarrow G^n$ given by projecting on the $n$ coordinates $i_1,\ldots i_n$. Let us denote by $N_{i_1,\ldots i_n}$ the kernel of this map. In general it need not be true that every finite index subgroup of $\Gamma$ contains one of the subgroups $N_{i_1,\ldots i_n}$. For example, if $G=\mathbb{Z}/p\mathbb{Z}$ where $p$ is a prime number, $\Gamma$ is a vector space over $\mathbb{Z}/p\mathbb{Z}$, and there will be many nontrivial homomorphisms $\psi:\Gamma\rightarrow \mathbb{Z}/p\mathbb{Z}$ such that $\psi(a_i)=0$ for every sequence $a_i$ with finite support, and therefore $Ker(\psi)$ will not contain any of the $N_{i_1,\ldots i_n}$ groups.

But what if we assume that $G=[G,G]$? Is it true now that every finite index subgroup contains $N_{i_1,\ldots i_n}$ for some sequence $i_1,\ldots,i_n$? More generally, is it true that all the $\mathbb{C}$-linear representations of $\Gamma$ (not just the continuous ones!) arise from pulling back representations of $G^n$ along $\phi_{i_1,\ldots i_n}$ for some sequence $i_1,\ldots i_n$?

• Regardless of $G$ you have, for any ultrafilter $\omega$, the kernel of the canonical homomorphism from $\prod G$ to the ultraproduct $\prod^\omega G$. If $\omega$ is non-principal the kernel contains the restricted product, hence is dense. – YCor Jun 10 '15 at 14:57
• You mean dense in the profinite topology? can one construct a finite quotient out of it? – Ehud Meir Jun 10 '15 at 15:06
• The ultraproduct $\prod^\omega G$ is finite. – YCor Jun 10 '15 at 16:17
• In fact the ultraproduct is isomorphic to $G$. This follows from Łoś's theorem since being isomorphic to a given finite group is a first-order property. – Qiaochu Yuan Jun 10 '15 at 23:51
• It is conceivable that any homomorphism to a finite group factors through to a finite product of ultraproducts. – Andreas Thom Jun 11 '15 at 4:09

Say that a tuple $f\in G^I$ vanishes at an ultrafilter $u\in \beta I$ if the support of $f$ is not in $u$. Let $N_{u_1,\ldots,u_n}$ be the (normal) subgroup of elements of $G^I$ that vanish at $u_1, \ldots, u_n$. In fact, if $X\subseteq \beta I$ let $N_X$ be the subgroup of elements that vanish at each $u\in X$. If $u_j$ is the principal ultrafilter consisting of sets containing element $i_j\in I$, then $N_{u_1,\dots,u_n} = N_{i_1,\dots,i_n}$, with the latter group in the OP's notation.

The OP's question was (essentially): if $G$ is perfect and $K\leq G^{\omega}$ is a subgroup of finite index, must $K$ contain some $N_{u_1,\ldots,u_n}$ where $u_1,\ldots, u_n$ are principal ultrafilters? YCor answered this negatively by noting that if $u$ is nonprincipal and $u_1,\ldots,u_n$ are principal, then $N_{u_1,\ldots,u_n}\not\subseteq N_u$

Andreas Thom, in his comment that begins "It is conceivable that $\ldots$", suggests a different form of the question. True or False: if $G$ is perfect and $K\leq G^I$ is a subgroup of finite index, must $K$ contain some $N_{u_1,\ldots,u_n}$ where $u_1,\ldots, u_n$ are not necessarily principal?

This version has an affirmative answer. To give an outline of the proof, let ${\mathfrak G} = G^I$ and define a mapping $\Gamma$ from the normal subgroup lattice of $\mathfrak G$ to itself: $\Gamma(A) = [{\mathfrak G},A]$ (commutator). This mapping is monotone ($A\subseteq B$ implies $\Gamma(A)\subseteq \Gamma(B)$) and decreasing ($\Gamma(A)\subseteq A$). If $A$ and $B$ are normal subgroups of $\mathfrak G$, write $A\subseteq_{\Gamma} B$ if there is some $k$ such that $\Gamma^k(A)\subseteq B$.

Claim 1. If $G$ is any finite group (not necessarily perfect) and $K$ is a subgroup of finite index in ${\mathfrak G} = G^I$, then there exist ultrafilters $u_1,\ldots,u_n$ such that $N_{u_1,\ldots,u_n}\subseteq_{\Gamma} K$.

Claim 2. If $G$ is perfect, then $\Gamma(N_{u_1,\ldots,u_n})=N_{u_1,\ldots,u_n}$. Hence $N_{u_1,\ldots,u_n}\subseteq_{\Gamma} K$ implies $N_{u_1,\ldots,u_n}\subseteq K$.

The Thom version of the question follows immediately from these two claims.

Idea for Claim 2: A product of groups is perfect iff each factor is. Hence if $G$ is perfect, so is $G^I$. If $G^I$ factors as $A\times B$, then $A$ and $B$ are perfect. This is enough to show that the kernel $N_C$ of the projection of $G^I\cong C(\beta I,G)$ onto a clopen set $V\subseteq \beta I$ is perfect. Now if $X\subseteq \beta I$ is arbitrary, then $N_X$ is the join of all $N_V$ where $X\subseteq V$ and $V$ is clopen, so $N_X$ is a join of perfect normal subgroups. This makes $N_X$ perfect for any subset $X\subseteq \beta I$. Now $N_X\supseteq \Gamma (N_X) = [{\mathfrak G}, N_X]\supseteq [N_X,N_X] = N_X$, proving $\Gamma(N_X)=N_X$.

Idea for Claim 1. If $A\lhd {\mathfrak G}$ is a meet-irreducible normal subgroup of finite index, then it has a unique cover $A^*$ in the normal subgroup lattice. Call the covering $A\prec A^*$ centralized if $[{\mathfrak G},A^*]\subseteq A$, else noncentralized. The proof is based on two observations:

(i) If $K\subseteq A\prec A^*$ are normal subgroups of finite index in $\mathfrak G$, where $A$ is meet-irreducible and $A\prec A^*$ is noncentralized, then there is an ultrafilter $u$ such that $N_u\subseteq A$.

(ii) If $L$ is a normal subgroup of $\mathfrak G$ that is contained in every normal subgroup $A\lhd {\mathfrak G}$ where: $K\subseteq A\prec A^*$, $A$ is meet-irreducible, $A\prec A^*$ is noncentralized, then there is a $k$ such that $\Gamma^k(L)\subseteq K$.

Here is how you put the observations together to prove Claim 1: Using (i), find and fix an ultrafilter $u$ such that $N_{u}\subseteq A$ for each noncentralized $A\prec A^*$ with $K\lhd A$. Form the intersection $N_{u_1,\ldots,u_n} = \cap N_{u_i} =:L$. This group is contained in every normal subgroup $A\lhd {\mathfrak G}$ where: $K\subseteq A\prec A^*$, $A$ is meet-irreducble, $A\prec A^*$ is noncentralized. By (ii), $N_{u_1,\ldots,u_n} \subseteq_{\Gamma} K$, which is what is needed to establish Claim 1.

Idea to prove (ii): $L$ is as described in (ii) iff $LK/K$ belongs to the hypercenter of ${\mathfrak G}/K$, hence $\Gamma^k(LK/K)$ is trivial for some $k$. For this $k$, $\Gamma^k(L)\subseteq K$.

Idea to prove (i): Suppose that $A\prec A^*$ are normal subgroups of finite index in ${\mathfrak G}=G^I$, where $A$ is meet-irreducible and $A\prec A^*$ is noncentralized. To find an ultrafilter $u$ such that $N_u\subseteq A$ it suffices to show that for each partition $I = Z\cup Z'$ of the index set into a subset and its complement it is the case that exactly one of the projection kernels $N_Z$ or $N_{Z'}$ is contained in $A$. Then $u=\{Z\subseteq I : N_Z\subseteq A\}$.

If $I=Z\cup Z'$, then it cannot be that both $N_Z\subseteq A$ and $N_{Z'}\subseteq A$, since $N_ZN_{Z'}={\mathfrak G} \not\subseteq A$.

To finish I must argue that if $I=Z\cup Z'$ is a partition, then at least one of $N_Z$ and $N_{Z'}$ is contained in $A$. This is accomplished by showing that if $N_Z\not\subseteq A\prec A^*$, then $[N_{Z'},A^*]\subseteq A$, i.e. $N_{Z'}$ centralizes $A\prec A^*$. If also $N_{Z'}\not\subseteq A$, then $N_{Z}$ centralizes $A\prec A^*$. But if they both centralize, then the join ${\mathfrak G} = N_ZN_{Z'}$ centralizes $A\prec A^*$, contrary to the choice of $A\prec A^*$.

So let's argue that if $N_Z\not\subseteq A$, then $[N_{Z'},A^*]\subseteq A$. Since $A$ is meet-irreducible in the normal subgroup lattice of $\mathfrak G$, with upper cover $A^*$, $N_Z\not\subseteq A$ implies $N_ZA = N_ZA^*$. By modularity, $N_Z\cap A\prec N_Z\cap A^*$. Moreover, a normal subgroup $M$ centralizes $A\prec A^*$ iff it centralizes $N_Z\cap A\prec N_Z\cap A^*$. But $N_{Z'}$ surely centralizes the latter, since $[N_{Z'},N_Z\cap A^*]\subseteq N_{Z'}\cap N_Z\cap A^* = \{1\}\subseteq N_Z\cap A$.