Need help in proving a subgroup result about the dyadic rationals

[michael.wes]

Homework Statement

Prove that every proper subgroup H of the dyadic rationals G (numbers of the form a/2^l, for integers a, l) which contains the integers is cyclic.

Homework Equations

The Attempt at a Solution

I was trying to argue based on the 'proper' requirement, i.e. there is some element in G - H of the form a/2^l. Then try and exhibit a generator based on this element. I then tried thinking about what the requirement that H must contain the integer means, and basically as far as I got was that a/2^l must be a certain kind of fraction so that you don't violate the requirement that the integers must be in H. But I'm still nowhere near finishing this. It;s one of the early problems on the assignment, so I might be overthinking this.

Any and all help is greatly appreciated!

Thanks,
M

 

[mighty2000]

aggie

Hello Maggie,

Thank you for your post. This is an interesting problem and I would be happy to help you with it. Let's start by defining some terms and concepts that will be helpful in solving this problem.

First, a proper subgroup H of a group G is a subgroup that is not equal to the entire group G. In other words, there are elements in G that are not in H.

Next, let's define the dyadic rationals G as the set of numbers of the form a/2^l, where a is an integer and l is a non-negative integer. This set includes all integers and fractions with powers of 2 in the denominator.

Now, let's think about what it means for a proper subgroup H of G to contain the integers. This means that all integers, which are of the form a/2^0, are in H. But since H is a subgroup, this also means that all other elements of the form a/2^l, where l>0, are also in H. This is because subgroups must be closed under the group operation, in this case, multiplication. So if we have an element a/2^l in H, we can multiply it by any other element in H, including the integers, and still get an element in H.

Now, let's consider your idea of using an element of the form a/2^l in G-H to generate a subgroup. This is a good start, but we need to make sure that this element is not already in H. Since H contains all the integers, we can't use an element with a non-zero denominator. This means we need to choose an element of the form 1/2^l, where l is a positive integer.

Now, let's think about what this element generates. Since H is a subgroup, it must contain the identity element, which is 1/2^0 = 1. So our element 1/2^l generates a subgroup of H that contains 1 and all other elements of the form 1/2^m, where m is a non-negative integer. This subgroup is cyclic, because it can be generated by a single element, 1/2^l.

But what about the elements of the form a/2^l, where a is an integer and l>0? These elements are not in the subgroup generated by 1/2^l, but