Dear All!


I tried for several evenings to find an answer to the following basic question and I cannot see what is the answer:


Given an integer $n\geq 3$, does there exist an (infinite) group with exactly $n$ normal subgroups?


If "yes", what about the same questions for finitely generated groups, finitely presented groups?


I guess this must have been done.



(F.g., f.p.) groups with exactly $n$ normal subgroups
If $n$ is even, then the answer is "yes". Take the direct product of a simple (infinite) group and ${\mathbb Z}/2^j{\mathbb Z}$. Every normal subgroup either is inside the finite cyclic group or contains the simple group. Total number is twice the number of normal subgroups of the cyclic group.If $n$ is odd, you would need to take a cyclic central extension of a simple group. That is also possible (the simple group can be, say, the Tarski monster, see our paper with Olshanskii and Osin on Lacunary hyperbolic groups in the arXiv).


 **Edit.**  If you want f.p. groups, look at the central extensions of the Thompson group $T$ described [here](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.80.628&rep=rep1&type=pdf).


