# 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.

• Please supply more context. – Derek Holt Mar 20 '15 at 19:39
• @DerekHolt iwill supply more context – user220373 Mar 21 '15 at 14:06
• I think the example you mention works. What is the problem? Are you having difficulty proving that it satisfies one of the required properties? – Derek Holt Mar 22 '15 at 14:08
• @DerekHolt yes exactly – user220373 Mar 22 '15 at 14:11
• So which property are you having difficulty proving? Surely it is clearly abelian by cyclic? – Derek Holt Mar 22 '15 at 16:02

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 $$G$$.
• I don't understand something in the last paragraph: $N$ is the normalizer of $H$, so $N$ contains $H$, and $N\cap H = H$. Then it is not cyclic. What is going on? – Ben Blum-Smith Jan 13 '19 at 22:26
• I meant $N \cap \langle a,b \rangle$. I have corrected this, as well as correcting the problem that $x$ unaccountably changed into $t$ halfway through the answer. – Derek Holt Jan 14 '19 at 8:52