Example of a Non-Abelian Infinite Group I was hunting an example of a non-trivial infinite group in which
1) All non-trivial normal subgroup are non-abelian.
2) There exists a nontrivial subnormal abelian subgroup.
Is there any hope to find this out?
Note
In the finite case such an example is not possible (see here).
Notation
A subgroup H of a given group G is a subnormal subgroup of G if there is a finite chain of subgroups of the group, each one normal in the next, beginning at H and ending at G.
 A: The following is a partial answer, it provides an example of a group $G$ that contains subnormal abelian subgroups inside some subgroup $M$ but no normal one inside this subgroup.  May be a variation of this yields a complete answer.   
Given a field $\mathbb{F}$, one can construct the free $\mathbb{F}$-module over the set of rational numbers.  Assume that $(v_x)_{x \in \mathbb{Q}}$ is a basis of this linear space (indexed by the rational numbers). For each pair of rational numbers $x <y$ consider $e_{xy}$ the linear transformation that maps $v_x$ to $v_y$, and maps the other elements of the basis to $0$. Now we define the McLain group $M=\operatorname{M}(\mathbb{Q},\mathbb{F})$ to be the group generated by all the linear transformations of the form $1+ae_{xy}$ , $x<y$ and $a \in \mathbb{F}$ (note that all of these transformations are invertible).   
One needs two facts about $M$, first, it is generated by abelian normal subgroups. Second, it is characteristically simple (that is it contains no proper trivial characteristic subgroup).  These facts are not very hard to prove, and one can find a proof of them in Robinson's "A course in the Theory of Groups".
Let $G$ be the semi-direct product of $M$ and $\operatorname{Aut}(M)$.
As $M$ contains non trivial abelian normal subgroups and $M$ is normal in $G$, $G$ contains subnormal non trivial abelian subgroups lying in $M$.  However, if $M$ contains an abelian $G$-invariant subgroup $A$, then $A$ is characteristic in $M$, a contradiction. 
One should notice that every normal abelian subgroup of $G$ intersects $M$ trivially, and centralizes $M$.
