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> Does there exist a group with a normal countable-index abelian subgroup but without characteristic countable-index abelian subgroups?
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It is well known that *any finite-index subgroup contains a normal (in the whole group) finite-index subgroup*. A lesser-known proposition states that


($\*$) *any group having an abelian finite-index subgroup has a characteristic abelian finite-index subgroup.*


(There are also a bunch of related not-so-easy facts [[Self-advertisement]](http://arxiv.org/abs/1309.0571)).


[K. Podoski and B. Szegedy](http://dx.doi.org/10.1515/jgth.2002.015) gave the best-known numerical version of ($\*$) (Theorem 5.1) and remarked that one cannot replace *finite* with *countable* in ($\*$); however, their example has no characteristic abelian countable-index subgroups just because it has no normal abelian countable-index subgroups. 



Large abelian characteristic subgroups in abelian-by-countable groups
Here is an example (the group $H\_K^\mathbf{Q}$ defined below, which is actually a split extension abelian-by-countable).


Let $(Q,\le)$ be a total ordering. Let $K$ be a countable (possibly finite) field. Let $V^Q\_K=K^{(Q)}$ be the free $K$-module with basis $(e\_q)\_{q\in Q}$. Let $G^Q\_K$ be the group of $K$-module automorphisms $g$ of $V\_K$ such that $g(e\_q)-e\_q\in\mathrm{Vect}\_Q(\{e\_r:r<q\})$ for every $q$ (it can be thought as a upper unipotent group of $Q\times Q$-matrices over $K$). 


Define a *cut* as a subset $I$ of $Q$ such that both $I$ and its complement $I^c$ are nonempty, and such that $i<j$ for all $i\in I$ and $j\in I^c$. For every cut $I$ with complement $J=I^c$ and $g\in G^Q\_K$, let $p\_I(g):V^I\_K\to V^I\_K$ be its restriction to the stable subspace $V^I\_K$, and let $r\_I(g):V^J\_K\to V^J\_K$ be the induced automorphism modulo the stable subspace $V^I\_K$. (Visually, $I$ defines a $2\times 2$ block decomposition, $p\_I(g)$ is the northwest block and $r\_I(g)$ is the southeast block of $g$.)


Obviously, $p\_I$ and $r\_I$ are homomorphisms, and if $P\_I$ and $R\_I$ are their kernels, then $N\_I=P\_I\cap R\_I$ is is a normal abelian subgroup in $G^Q\_K$.


Now we specify to $Q=\mathbf{Q}$, the rationals with the usual ordering. 
Still, $N\_I$ is not of countable index in $G^\mathbf{Q}\_K$. So we define $H^Q\_K$ as the normal subgroup of $G^Q\_K$ of elements $g$ such that for every cut $I$, both matrices $p\_I(g)$ and $r\_I(g)$ are finitely supported (i.e. finitely many non-diagonal elements are nonzero). Then $N\_I\cap H^\mathbf{Q}\_K$ is normal abelian of countable index in $H\_K^\mathbf{Q}$.


It remains to see that $H\_K^\mathbf{Q}$ has no abelian characteristic subgroup of countable index. This is rather easy: the first observation (an easy play with commutators(\*)) is that every nontrivial normal subgroup $N$ of $H\_K^\mathbf{Q}$ contains an elementary matrix $e\_{ij}(s)$ for some $i<j$ and $s\neq 0$. All such matrices are conjugate under the full automorphism group of $H\_K^\mathbf{Q}$ (to vary $s$, use diagonal matrices, and to vary $(i,j)$, use that the group of order-preserving permutations of $\mathbf{Q}$ is 2-transitive). Hence if $N$ is characteristic, it contains all elementary matrices, and thus the (non-abelian) subgroup they generate.


Hence the only abelian characteristic subgroup of $H\_K^\mathbf{Q}$ is the trivial subgroup $\{1\}$: we have to check that it does not have countable index, i.e. $H\_K^\mathbf{Q}$ is uncountable: indeed for every subset $L$ of $\mathbf{N}\smallsetminus\{0\}$, consider the matrix $k\_L$ with same entries as the identity except $(k\_L)\_{-i,i}=1$ for all $i\in L$. Then $L\mapsto k\_L$ is injective, $k\_L$ belongs to $H\_K^\mathbf{Q}$ for every $L$, hence $H\_K^\mathbf{Q}$ has continuum cardinal.


Edits: 


(1) (\*) Starting from a non-identity element $g$ in $G\_K^\mathbf{Q}$, its commutator with a suitable elementary matrix yields a non-identity element $h$ such that $h-1$ has finite rank; in particular if $g\in H\_K^\mathbf{Q}$ then so does $h$, and it follows that $h$ is finitely supported. Then $h$ lies in the nilpotent subgroup $G\_K^I$ for some finite subset $I$ of $\mathbf{Q}$. So the normal subgroup of $H\_K^\mathbf{Q}$ generated by $g$ contains a central element of $G\_K^I$, showing that any nontrivial normal subgroup of $H\_K^\mathbf{Q}$ contains a non-identity elementary matrix.


(2) Actually, a similar argument shows that the only normal abelian subgroup of $H\_K^\mathbf{Z}$ stable under the shift automorphism $\alpha$ (conjugation by the automorphism $(e\_n\mapsto e\_{n+1})\_{n\in\mathbf{Z}}$) is the trivial subgroup $\{1\}$. Note that $H\_K^\mathbf{Z}$ is residually nilpotent, while $H\_K^\mathbf{Q}$ is a perfect group.


(3) Incidentally, with the notation of (2), the semidirect product $\langle\alpha\rangle\ltimes H\_K^\mathbf{Z}$ has a countable index abelian subgroup but no normal one. I don't know whether the Podoski-Szegedy example is similar (I don't have access to their article right now).


