finding 3 groups that $G_1 \subsetneq G_2 \subsetneq G_3$ such that $G_1 \lhd G_2$ but $G_2 \not\lhd G_3$

Question

Find three groups $G_1, G_2, G_3$ such that $({\text{id}}, ◦) \subsetneq G_1 \subsetneq G_2 \subsetneq G_3$ (that is, each is properly included in its following) such that $G_1 \lhd G_2$ but $G_2 \not\lhd G_3$.

That is the question I have been asked, I have tried to find these three groups in non abelian groups such as $D_n$ or $S_n$ but couldn't find the right conditions ( for example in $D_3$ i found a normal subgroup $\{e,(123),(132)\}$ and a non normal subgroup $\{e,(12)\}$ but they are not included on in another).

Please someone can guide me how to figure this out ?

It's common notation for triangle between two subgroups to mean normal subgroup, yes. What does the square symbol mean between two subgroups? Oh, just noticed yu now use triangle in both. I'll delete comment soon. — coffeemath, Jun 08 '20 at 13:59
I suspect that your triangles were backwards. Should this not be the case, please reverse my edit. — AnalysisStudent0414, Jun 08 '20 at 14:12

score 1 · Accepted Answer · 2020-06-08T14:54:37.957

1

You may take $G_3=S_4$, $G_2=S_3$ (the subgroup fixing $4$) and $G_1=A_3$. Note that $S_4$ cannot have a normal subgroup of order 6, and thus $G_2$ is not normal in $G_3$. Hopefully, you already know that $A_3$ is normal in $S_3$.

edited Jun 08 '20 at 14:54

answered Jun 08 '20 at 14:38

finding 3 groups that $G_1 \subsetneq G_2 \subsetneq G_3$ such that $G_1 \lhd G_2$ but $G_2 \not\lhd G_3$

1 Answers1