2
$\begingroup$

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 $i_1,\ldots i_n$?

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – YCor Jun 10 '15 at 14:57
  • $\begingroup$ You mean dense in the profinite topology? can one construct a finite quotient out of it? $\endgroup$ – Ehud Meir Jun 10 '15 at 15:06
  • 1
    $\begingroup$ The ultraproduct $\prod^\omega G$ is finite. $\endgroup$ – YCor Jun 10 '15 at 16:17
  • $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Jun 10 '15 at 23:51
  • 1
    $\begingroup$ It is conceivable that any homomorphism to a finite group factors through to a finite product of ultraproducts. $\endgroup$ – Andreas Thom Jun 11 '15 at 4:09
1
$\begingroup$

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 $[N_{Z'},N_Z\cap A^*]\subseteq N_{Z'}\cap N_Z\cap A^* = \{1\}\subseteq N_Z\cap A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.