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It is well-known that each abelian group admits an injective homomorphism to some compact topological group (for example to its Bohr compactification). Is the same fact true for solvable groups?

Question 1. Does every solvable group admit an injective homomorphism to a compact Hausdorff topological group?

Equivalently:

Question 2. Is the Bohr topology of every discrete solvable group Hausdorff?

The Bohr topology is the largest totally bounded group topology on the group.

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    $\begingroup$ No. A finitely generated group admits an injective homomorphism to a compact group iff it's residually finite. There exist finitely generated solvable groups that are not residually finite. $\endgroup$
    – YCor
    Commented May 3, 2017 at 23:58
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    $\begingroup$ @YCor: Is there a reference (or short explanation) for this criterion? The "if" direction is clear, but I don't immediately see why the converse holds. $\endgroup$
    – Victor Protsak
    Commented May 4, 2017 at 4:48
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    $\begingroup$ I think this should work: compact (Hausdorff) groups are pro-(matrix Lie groups) by the Peter-Weyl theorem, and finitely generated subgroups of pro-(matrix Lie groups) are residually finite by Malcev's theorem. $\endgroup$
    – Qiaochu Yuan
    Commented May 4, 2017 at 7:46
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    $\begingroup$ @VictorProtsak Qiaochu's right: Peter-Weyl + Malcev. For the existence of such solvable groups: two such families appeared in the 50's: f.g. groups solvable whose center is isomorphic to the non-residually-finite abelian group $\mathbf{Z}[1/p]/\mathbf{Z}$ (Ph. Hall), or Gruenberg, who proved that every wreath product $H\wr\mathbf{Z}$ for $H$ an arbitrary non-abelian group, is non-residually-finite. $\endgroup$
    – YCor
    Commented May 4, 2017 at 8:48
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    $\begingroup$ f.g. finite-by-abelian groups are virtually abelian (= abelian-by-finite) and hence residually finite. More generally virtually nilpotent f.g. groups are residually finite. This is well documented (certainly in Derek Robinson's Springer GTM book). $\endgroup$
    – YCor
    Commented May 4, 2017 at 11:50

1 Answer 1

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According to Proposition 3.3 from Dikranjan and Toller [Topology and its Applications 159 (2012) 2951-2972], the Heisenberg group H_K over an infinite field K of characteritic 0 is not maximally almost periodic. This provides a negative answer to the question even for nilpotent groups of class 2.

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    $\begingroup$ The char 0 case was well-known before. Indeed, consider a homomorphism of $H_3(K)$ into a compact Lie group. Since $H_3(K)$ has no finite proper quotient as abstract group, the closure $G$ of the image is connected; as a connected nilpotent compact Lie group, it is abelian. So by Peter-Weyl, every homomorphism from $H_3(K)$ to a compact group has abelian image. $\endgroup$
    – YCor
    Commented Jun 18, 2017 at 8:48
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    $\begingroup$ The char $p$ case, possibly well-known too, is by a distinct argument. First, we see that since a compact Lie group of bounded exponent is finite and using Peter-Weyl, a group of finite exponent is maximally almost periodic if and only if it's residually finite. Then whether these kinds of 2-nilpotent groups are residually finite has probably been considered early; anyway it's less obvious than in char. 0. $\endgroup$
    – YCor
    Commented Jun 18, 2017 at 9:02
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    $\begingroup$ Here's the argument that $H=H_3(K)$ is not RF (and hence not MAP) in char $p$, namely every finite quotient is abelian. Otherwise there exists a finite index subgroup $N$, intersecting the center $Z$ in a proper subgroup $M$. Enlarging $M$, we can suppose that $M$ has index $p$ in $Z$. For a contradiction, let's show that $Z/M$ has abelian image in every finite quotient of $H/M$. The commutator (valued in $Z/M$) yields an alternating $Z/pZ$-bilinear form $b$ on $V=H/Z$. This form has trivial kernel, because in $H$ the map $y\mapsto [x,y]$ is surjective for any fixed $x\neq 0$ in $V$. (...) $\endgroup$
    – YCor
    Commented Jun 18, 2017 at 11:40
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    $\begingroup$ Hence there exist infinite sequences $(x_n)$, $(y_n)$ in $V$ such that $b(x_n,y_n)\neq 0$ (in $H/M$) while $b(x_n,x_m)=0$ for all $n,m$ and $b(x_n,y_m)=0$ for all $n\neq m$. Inside any finite quotient $F$ of $H/M$ in which $Z/M$ is preserved, these properties still hold. But on the other hand, they imply that in the quotient $F/(Z/M)$ of $V$, the family $(x_n)$ is linearly free (over $\mathbf{Z}/p\mathbf{Z}$). It contradicts that $F$ is finite. $\endgroup$
    – YCor
    Commented Jun 18, 2017 at 11:45

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