From: Jan Kristian Haugland Subject: Re: Subgroups of Z^n Date: Sun, 31 Dec 2000 14:00:50 +0100 Newsgroups: sci.math Summary: [missing] Posted on behalf of Avinoam Mann ------------------------------------------ Date: Sun, 31 Dec 2000 13:24:24 +0200 (IST) From: AVINOAM MANN Subject: Re: Subgroups of Z^n To: noam_katz@my-deja.com, israel@math.ubc.ca, jkhaug00@stud.hia.no cc: mann@vms.huji.ac.il > noam_katz@my-deja.com wrote: > > > Let G be the group Z^d (d times direct sum of the integers Z). > > If m >= 2 how many subgroups of index m there are in G ? > Because of some problem with my news reader, I cannot post directly to sci.math, so I write to the people who showed interest in that question. You may post this if you wish. Let a_m be the number in question, and let f(s) be the Dirichlet series $\sum a_m/m^s$. Let Z(s) be Riemann's zeta function. Then f(s) is the reciprocal of the product Z(s)Z(s-1)...Z(s-d+1). E.g. if d = 2 we see that a_m is the sum of the divisors of m. For a proof see, e.g., A.Lubotzky's paper _Counting subgroups of finite index_ in v.2 of the Proceedings of the St. Andrews/Galway 93 group theory meeting, Th. 2.1. This v.2 is no. 212 in the LMS Lecture Notes Series, Cambridge University Press 1995. Lubotzky gives references to four other proofs, none of them difficult. Best wishes for the new millenium Avinoam Mann ============================================================================== From: israel@math.ubc.ca (Robert Israel) Subject: Re: Subgroups of Z^n Date: 29 Dec 2000 21:51:53 GMT Newsgroups: sci.math In article <92htij$5jg$1@nnrp1.deja.com>, wrote: >Let G be the group Z^n (n times direct sum of the integers Z). >If m >= 2 how many subgroups of index m there are in G ? If m is prime, the answer is (m^n-1)/(m-1). This can be shown by induction on n. For n=1 it is trivial. If there are f(n-1) subgroups of index m in Z^(n-1), then in Z^n there are the f(n-1) subgroups Z x H, where H is a subgroup of index m in Z^(n-1), plus the m^(n-1) subgroups of the form {x: x_1 + sum_{i=2}^n b_i x_i = 0 mod m} where b_2, ..., b_n are in {0,...,m-1}. It is not hard to show that every subgroup of index m must be one of these. So f(n) = f(n-1)+m^(n-1) (which is a recursion satisfied by f(n) = (m^n-1)/(m-1)). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2