# 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.

• 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-Álvarez 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.