Question about self homomorphism of dyadic rationals

Question

I'm reading a paper that uses the group $D = \{j/2^n \mid j \in \mathbb{Z}, n \geq 0\}$ and they define a homomorphism $h: D \rightarrow D$ determined by $h(1) = j/2^n$ for some odd $j$ and positive integer $n$. My question is, how can one determine $h(m/2^j)$ for any other $m/2^j \in D$. I don't believe one can treat it as a ring homomorphism since I already tried doing that and arrived at getting something not in $D$. Also, they say that the image of $h$ is a subgroup of odd index. Why is that? Sorry if this question seems kind of trivial, my group theory knowledge and skills are lacking and I'd rather ask someone who knows rather than pour through books at the moment.

Answer 1

$h(1)$ being determined, $h(n)$ is clearly determined fo $n\in \mathbb{Z}$.

The point is now that multiplication in $\mathbb{Z}$ is special : $n\times m = m +....+m$ where there are $n$ $m$'s (if $n\geq 0$). So, say you want to determine $h(\frac{j}{2^n})$. Then look at $h(j) = h(2^n \frac{j}{2^n}) = h(\frac{j}{2^n} +...+\frac{j}{2^n}) = h(\frac{j}{2^n}) +...+h(\frac{j}{2^n}) = 2^n h(\frac{j}{2^n})$.

Now in $D$, for all $a$, the equation (unknown $x$) $2^n x = a$ has exactly one solution, and that is $x=\frac{a}{2^n}$.

So in the end we get $h(\frac{j}{2^n}) = \frac{jh(1)}{2^n}$. That's why the value of $h(1)$ completely determines $h$