# Large abelian characteristic subgroups in abelian-by-countable groups

Does there exist a group with a normal countable-index abelian subgroup but without characteristic countable-index abelian subgroups?

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]).

K. Podoski and B. Szegedy 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.

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).

• Thank you, Yves!! Surely, you mean $g(e_q)\in e_q+Vect(\dots)$ in the first paragraph. Also, $G_K$ in the second paragraph is the same as $G_K^Q$. Furthermore, I guess that the "easy play with commutators" is something like the following. The commutator of any matrix and an elementary matrix is finitely supported, i.e. this commutator lies in a finite-dimensional unitriangular subgroup $UT_n(K)$ which is nilpotent and, hence, any its non-trivial normal subgroup non-trivially intersects the centre, which consists of elementary matrices. – Anton Klyachko Dec 14 '13 at 22:36
• @AntonKlyachko I edited the typos. For the commutators I had a similar argument in mind (although yours is more elegantly written than what I had in mind). Note that "the commutator of any matrix and an elementary matrix is finitely supported" holds in $H$ but not in $G$. – YCor Dec 14 '13 at 22:44
• Excuse me, Yves. I do not understand your comment. The finitely supported matrices form a normal subgroup of $G$, right? If they do, then the commutator must be finitely supported. – Anton Klyachko Dec 15 '13 at 0:06
• @Anton: I don't see what you mean: if you consider the matrix $g$ with same entries as identity except $g_{1j}=1$ for all $j>1$, then $[e_{01},g]_{0j}=1$ for all $j>1$. (Note that $G$ is the set of upper unipotent matrices for which all columns are finitely supported, though lines can be infinitely supported.) – YCor Dec 15 '13 at 0:41
• Oh, yes. I mixed up finitely supported matrices and matrices $f$ with finite $\hbox{rank}(f-1)$ (i.e operators $f$ such that $\hbox{ker}(f-1)$ is of finite codimension). The latter form a normal subgroup but the former do not. Inside $H$, these two subgroups coincides, right? (I mean, their intersections with $H$ coincides.) – Anton Klyachko Dec 15 '13 at 2:03