4
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ How many subgroups does a finite cyclic group have? $\endgroup$
    – HJRW
    Commented Aug 24, 2011 at 15:16

2 Answers 2

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ The next variation might combine this with the requirement that the group be not residually less than kappa for certain cardinals kappa. Gerhard "Ask Me About System Design" Paseman, 2011.08.24 $\endgroup$
    – Gerhard Paseman
    Commented Aug 24, 2011 at 16:09
  • $\begingroup$ A simpler 'next' variant would be to ask for a torsion-free example. $\endgroup$
    – HJRW
    Commented Aug 25, 2011 at 10:49
  • 1
    $\begingroup$ @HW: Such a group can be constructed for $n$ powers of 2 as direct powers of the Burger-Mozes group $G$ (or any other torsion-free simple group). The group $G^n$ has exactly $2^n$ normal subgroups. If $n$ is not a power of 2 one needs to build extensions towers of simple groups. That's not going to be easy. $\endgroup$
    – user6976
    Commented Aug 25, 2011 at 21:17
3
$\begingroup$

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."

Improve this answer
$\endgroup$
3
  • $\begingroup$ Sorry, I meant infinite groups of course! $\endgroup$
    – Victor
    Commented Aug 24, 2011 at 15:31
  • $\begingroup$ Surely Daniel's answer deserves to be accepted? It answers the question you asked. $\endgroup$
    – HJRW
    Commented Aug 25, 2011 at 10:48
  • $\begingroup$ @HW: That's alright; it wasn't the question he intended to ask. I certainly don't mind. $\endgroup$
    – Daniel Litt
    Commented Aug 25, 2011 at 15:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .