# Weird quotient of $\langle\mathbb Q,+\rangle$?

after looking at this question I came to think on one particular case. I'm wondering if maybe I've missed something on the way. If anyone could give it a look that would be great:

We start by considering a subset $H=\lbrace\frac{p}{q}\in\mathbb Q| p\text{ is even and }q\text{ is odd}\rbrace$.

Naturally $0=\frac{0}{1}\in\mathbb Q$, and for any $\frac{p}{q}\in H$: $\frac{-p}{q}\in H$. Also, if $\frac{p_1}{q_1},\frac{p_2}{q_2}\in H$ then $p_1q_2+p_2q_1$ is even, and $q_1q_2$ is odd, so $H$ is indeed a sub-group of $\mathbb Q$ (sorry for going into too much detail, as I said- I'm trying to make sure I didn't miss anything)

So- what's $\mathbb{Q}/H$? Let's see what we've got: Let $r_1,r_2\in\mathbb{Q}$ be arbitrary. Write them down as $r_1=2^{t_1}\frac{p_1}{q_1}$ and $r_2=2^{t_2}\frac{p_2}{q_2}$ with $p_1,p_2,q_1,q_2$ all odd integers, and $t_1,t_2\in\mathbb{Z}$

We know that $$r_1=r_2\mod H\iff r_1-r_2\in H\iff \frac{2^{t_1}p_1q_2-2^{t_2}p_2q_1}{q_1q_2}\in H$$ WLOG assum $t_1\geq t_2$ then we have $$2^{t_2}\frac{2^{t_1-t_2}p_1q_2-p_2q_1}{q_1q_1}\in H$$ From here on I notice two cases:

Case 1: If $t_2>0$ than this is always true- meaning that any two elements of $\mathbb{Q}$ with a positive diadic valuation are congruent under $H$ (since $t_1\geq t_2>0$). EDIT- Not very suprising, as any element of positive diadic valuation is $0_{\mathbb Q/H}$, by definition (Thanks to Brian for mentioning this).

Case 2: (and this is what's got me baffled) If $t_2\leq 0$ we need $2^{t_1-t_2}p_1q_2-p_2q_1$ to be an even integer, such that $2^{-t_2}$ divides it, for this number to be in $H$. Since $p_2q_1$ is odd, it must hold that $2^{t_1-t_2}=1$ so $t_1=t_2$. Also we have that $2^{-t_2}|p_1q_2-p_2q_1$.

Basically this is where I got stuck- what's the deal with Case 2? What kind of a group is this? Am I completely off track somewhere? or is this maybe a known result?

If anyone can point me at some direction here I would be very thankful.

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I'm not sure if the p-adic tag is relevant. Anyone who thinks otherwise, please edit :) –  kneidell Jul 5 '11 at 20:03
If $t_2 > 0$, $r_1,r_2 \in H$, so Case 1 is just picking out $0_{\mathbb Q/H}$, isn't it? –  Brian M. Scott Jul 5 '11 at 20:23
And in Case 2 you want $2^{-t_2} | p_1q_2 - p_2q_1$. –  Brian M. Scott Jul 5 '11 at 20:30
@Brian: yes, you're right, I didn't see it- Case 1 is just picking $0$. about $2^{-t_2}$, i've change it. –  kneidell Jul 5 '11 at 20:38
What kind of answer do you want to «what's $\mathbb Q/H$?»? It can be described as $\operatorname{colim}_{q\text{ odd}}\mathbb Z_{2q}$ with the colimit taken over odd numbers $q$ ordered by divisibility (I think). Does that qualify as an answer? –  Mariano Suárez-Alvarez Jul 5 '11 at 20:48

Let $G = \mathbb Q/H$. For $x,y \in \mathbb Q$ write $x \sim y$ if $x-n \in H$, and write $[x]$ for the $\sim$-equivalence class of $x$.
$H = \{x \in \mathbb Q:|x|_2 < 0\}$. For $n \in \omega$ let $A_n = \{x \in \mathbb Q:|x|_2 = n\}$.
Claim: If $x \in A_n$, there is an odd integer $m$ s.t. $0 < m < 2^{n+1}$ and $x \sim \frac{m}{2^n}$.
To see this, write $x = \frac{a}{2^nb}$, where $a$ and $b$ are odd; there are odd integers $q,r$ s.t. $a = 2^{n+1}q + r$ with $0<r<2^{n+1}$. Since $b$ and $r$ belong to the multiplicative group mod $2^{n+1}$, there is an odd integer $m$ s.t. $0 < m < 2^{n+1}$ and $r-bm = 2^{n+1}k$ for some integer $k$. Then $$x - \frac{m}{2^n} = \frac{2^{n+1}q + r}{2^nb} - \frac{m}{2^n} = \frac{2^{n+1}q+r-bm}{2^nb} = \frac{2(q-k)}{b} \in H,$$ and $x \sim \frac{m}{2^n}$.
On the other hand, if $0 < k < m < 2^{n+1}$, and $k$ and $m$ are odd, $\frac{m}{2^n} - \frac{k}{2^n} = \frac{m-k}{2^n} \notin H$, since the highest power of $2$ that can divide $m-k$ is $2^n$. It follows that $A_n/H = \{[\frac{2k+1}{2^n}]:0 \le k < 2^n\}$, where the $2^n$ elements listed are all distinct. It's also clear that the elements of $A_n/H$ are precisely the elements of order $2^n$ in $G$. I'm no algebraist and have never dealt with the $2$-Prüfer group before, but if I understand the Wikipedia description correctly, Jyrki is correct, and that's what we have here.