Re: group problem

Li Yi <liyi.cn@xxxxxxxxx> wrote:

> So how to prove it?

Let's see ....

Let o(a) = m and o(b) = n.

Let m1 be the smallest positive integer such that a^m1 is a power of b,
which we will call b^n2. Let d1 = m/m1 and n1 = (n,n2). It is easy to
show that b^n1 is a power of a and that n1*d1 = n. Also, let d2 =
(m1,n1) and note that d1*d2 = (m,n).

Let H be the set of group elements that can be written as a^k b^l, where
0 <= k < m1 and 0 <= l < n. Note that all m1*n expressions are distinct
elements, and that H is a subgroup of G. Consider the set of elements
h^m1. This is a subgroup of the group generated by b. Since it
contains b^n1 and b^m1, it must contain b^d2; in fact it is generated by
b^d2. Let H_x be the subset of H whose elements h satisfy h^m1 = b^x.
Note that multiplying any element of H_x1 with an element of H_x2 yields
an element of H_(x1+x2). This implies that H_(x+d2) is at least as
large as H_x, since choosing one element of H_d2 for that product yields
an appropriate number of distinct elements of H_(x+d2). The size of
each nonempty H_x must therefore be m1*d2.

The set H_d2 must include at least one element of the form a^k b^l where
k is coprime with m1; otherwise, there is no way for a to be in any of
the H_x sets. Consider the order of this element. It must be a
multiple of m1, or else that power isn't even a power of b. And if it
is m1*n3, where n3 < n/d2, that power becomes b^(d2*n3), which is not e.
But m1*(n/d2) obviously works as an exponent, so that must be the order.

m1*(n/d2) = (m/d1)*(n/d2) = (m*n)/(d1*d2) = (m*n)/(m,n) = [m,n]
--
Daniel W. Johnson
panoptes@xxxxxxxxxx
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
.

Relevant Pages

  • Re: Matrix Algebra question
    ... the statment is not correct. ... That correspondence is not one-to-one. ... Now multiplying by a power of L gets you to I or -I. ...
    (sci.math)
  • Re: Matrix Algebra question
    ... Now multiplying by a power of L gets you to I or -I. ... So what is the correct characterization of G then? ... In fact the diagonal entries of G are easily seen to be congruent to 1 ...
    (sci.math)
  • Re: Matrix Algebra question
    ... I am looking for an easy proof of the following fact: ... Now multiplying by a power of L gets you to I or -I. ... Rudin was correct since A and -A correspond to same linear ...
    (sci.math)
  • Re: Tetration again
    ... to rationals is an interesting generalization seems to be ... number of times" or "multiplying a number a fractional number of ... integer or raising to an integer power. ...
    (sci.math)
  • Re: Matrix Algebra question
    ... I am looking for an easy proof of the following fact: ... Now multiplying by a power of L gets you to I or -I. ... So what is the correct characterization of G then? ...
    (sci.math)