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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.

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How many subgroups does a finite cyclic group have? –  HJRW Aug 24 '11 at 15:16
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2 Answers

up vote 8 down vote accepted

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.

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That's great! Thanks a lot!!! –  Victor Aug 24 '11 at 15:53
    
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 –  Gerhard Paseman Aug 24 '11 at 16:09
    
A simpler 'next' variant would be to ask for a torsion-free example. –  HJRW Aug 25 '11 at 10:49
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@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. –  Mark Sapir Aug 25 '11 at 21:17
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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."

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Sorry, I meant infinite groups of course! –  Victor Aug 24 '11 at 15:31
    
Surely Daniel's answer deserves to be accepted? It answers the question you asked. –  HJRW Aug 25 '11 at 10:48
    
@HW: That's alright; it wasn't the question he intended to ask. I certainly don't mind. –  Daniel Litt Aug 25 '11 at 15:20
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