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.

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

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  • $\begingroup$ 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? $\endgroup$ – Ben Blum-Smith Jan 13 '19 at 22:26
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Jan 14 '19 at 8:52

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