Example of three-generator abélien by cyclic Can you give me example of three-generator group abelien by cyclic(i.e there exist normal subgroup $N$ in $G$ abelien and $G/N$ is cyclic) which is not finite by nilpotent (i.e there isn't  finite subgroup$ M$ in $G$ such that   $G/M$ is nilpotent)  but has each of its two-generator subnormal subgroups abelien.
for example let $A$ be free abelien group of rank $2$ generated by $a$,$b$ and let $x$ be the automorphism of order 2 which invert evry element of $A$ ,let $G$ be split extension of A by$<x>$  .
subnormal subgroups if There is a finite ascending chain of subgroups starting from the subgroup and going till the whole group.

We want to prove that every subnormal subgroup of the group $G=\langle a,b,t \mid ab=ba, t^2=1, a^t=a^{-1},b^t=b^{-1} \rangle$ is abelian.
Let $H$ be a nonabelian $2$-generator subgroup of $G$. Then $H$ contains some element outside of $\langle a,b \rangle$, and since all such elements have order $2$ and are equivalent under an automorphism of $G$, we can assume that $t \in H$. We can assume also that the other generator lies in $\langle a,b \rangle$, so $H = \langle t,a^ib^j \rangle$ for some $i,j \in {\mathbb Z}$. 
We need to show that $H$ is not subnormal in $G$. Let $N = N_G(H)$. If $a^kb^l \in N$, then $a^{-k}b^{-l}ta^kb^l = t a^{2k}b^{2l} \in H$, so $a^{2k}b^{2l} \in H$. Hence $|N:H| \le 2$ and $N  \cap \langle a,b \rangle$ is still cyclic. So, if $|N:H| = 2$ we can replace $H$ by $N$ and calculate its normalizer. Repeating this process, the ascending chain of normalizers must eventually stabilize with a $2$-generator subgroup $N'$ with $N' \cap \langle a,b \rangle$ cyclic. So $H$ is not subnormal in $G$.
