Infinite group with no maximal normal solvable subgroup

If $G$ is a finite group, then it has a maximal solvable normal subgroup. I think that this fails for infinite groups, but I don't know an example. Can you name one?

• Are you looking for that example in non-abelian groups? – mrs Dec 1 '13 at 16:29
• If the group were abelian, it would be its own, unique, maximal, normal, solvable subgroup - so I am not expecting a counterexample there. More precisely, the group must certainly not be solvable. – Jesko Hüttenhain Dec 1 '13 at 16:46
• In fact, I know that if $G$ is divisible so it has no any maximal subgroup. Thanks – mrs Dec 1 '13 at 16:50
• I don't understand your last comment. Note that I do not require the subgroup to be proper; if $G$ is solvable, it satisfies all conditions because $G$ is a subgroup of itself. It is also clearly maximal with these properties. – Jesko Hüttenhain Dec 1 '13 at 17:16
• Take the union of any infinite ascending chain of finite soluble groups of unbounded derived length. For example, start with a cyclic group, and keep taking wreath products and embedding each one in the next. – Derek Holt Dec 1 '13 at 20:42

Take the group $U$ of upper triangular infinite (in the bottom-right direction) matrices with real coefficients (or coefficients from your favorite commutative ring), with all $1$'s on diagonal so that all but finitely many off-diagonal values are zero: $$\left[\begin{array}{cccc} 1 & a_{12} & a_{13} & \ldots\\ 0 & 1 & a_{23} & \ldots \\ \vdots & \vdots & \vdots & \vdots \\ \end{array}\right]$$ This will give you an example: To see an increasing chain of normal solvable subgroups (of derived length increasing to infinity), consider first the abelian subgroup $N_1<U$, where the only nonzero off-diagonal entries occur in the 1st row. It is easy to see that $N_1$ is normal in $U$. Next, conider the subgroup $N_2$ consisting of matrices where the only nonzero off-diagonal entries occur in the first two rows, and so on. The subgroup $N_n$ thus constructed will have derived length $n$ and $N_n$ is normal in $U$. It is not hard to see that $U$ will not have a maximal solvable subgroup.