(F.g., f.p.) groups with exactly $n$ normal subgroups - MathOverflow

(F.g., f.p.) groups with exactly $n$ normal subgroups

2011-08-24T15:00:46Z

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.

Answer by Daniel Litt for (F.g., f.p.) groups with exactly $n$ normal subgroups

2011-08-24T15:16:18Z

In general, $\mathbb{Z}/2^k\mathbb{Z}$ has $k+1$ normal subgroups, namely $\mathbb{Z}/2^j\mathbb{Z}$ for $0\leq j\leq k$. So the answer to your question is "yes."

Answer by Mark Sapir for (F.g., f.p.) groups with exactly $n$ normal subgroups

2011-08-24T15:50:25Z

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.