# What does $\Bbb Z[\frac16]/\Bbb Z$ and its subgroups look like?

What does $\Bbb Z[\frac16]/\Bbb Z$ and its subgroups look like? In particular, I'm trying to get a grip on how many elements there are of any given order, and what their graph looks like.

I'm referring to the additive group with addition modulo $1$, where $\Bbb Z[\frac16]$ is the subfield of $\Bbb Q$

So I'm thinking of these as the fractions in the interval $[0,1)$ having denominators drawn from $2^n\cdot3^m$.

• In the case of $\Bbb Z[\frac12]/\Bbb Z$ it's fairly straightforward as there are $2^{n-1}$ elements of order $2^n$, those being the odd integers over $2^n$ in the interval.

• The subgroups of $\Bbb Z[\frac12]/\Bbb Z$ are also fairly trivial to describe as being the Prufer 2-group the hierarchy is a one-dimensional descent through the powers $2^n$

• The Prufer 2-group can easily be graphed by the regular, infinite, binary rooted tree.

What are the comparable three statements for $\Bbb Z[\frac16]/\Bbb Z$? I presume a little complexity is introduced by having a combination of two prime factors in the denominators.

• The elements of order $q$ are the elements of the form $p/q$ with $\gcd(p,q)=1$. – Gerry Myerson Jul 10 '18 at 9:32
• It is just the direct sum of the Prufer 2-group with the Prufer 3-group. – Derek Holt Jul 10 '18 at 9:39
• The group operation is addition, not multiplication. – Derek Holt Jul 10 '18 at 11:29
• Yes that defines an isomorphism. In general any abelian torsion group is the direct sum of its primary components. See math.stackexchange.com/questions/22646 – Derek Holt Jul 10 '18 at 12:28
• In particular, to find $y$ and $z$ one has to solve an equation of the form $a2^m+b3^n=c$; since $\gcd(2^m,3^n)=1$, one can find a 'fundamental solution' of $i2^m+j3^n=1$ using the Euclidean algorithm and then take $a=i\cdot c, b=j\cdot c$. – Steven Stadnicki Jul 10 '18 at 15:26