Quotient groups of the positive real numbers

Question

I'm having trouble working with the definition of a quotient group. In particular I'm trying to find certain quotient groups $\mathbb{R}^+/P$, where $P$ is a subgroup of $(\mathbb{R}^+,\cdot)$. I looked at the most simple non-trivial subgroups, i.e. $P=\{a^n:n\in\mathbb{Z}\}$ for $a\ne 1$. Isn't it the case that for every $x\in\mathbb{R}^+$ which is not a power of $a$, the coset $xP$ contains the element $x$ and thus the quotient group just ends up containing all the positive real numbers? Also can one find a homomorphism on $\mathbb{R}^+$ with $\text{ker}f=P$ such that by the fundamental theorem of homomorphisms the quotient group $\mathbb{R}^+/P=\text{ran}f$?

EDIT: I think, I almost got it. Let w.l.o.g. $a>1$. So, for every $x\in\mathbb{R}^+$ you can choose some $n\in\mathbb{Z}$ such that $xa^n\in[1,a)$ (since there exists unique $n\in\mathbb{Z}$, s.t. $x\in[a^{-n},a^{-n+1})$), hence the set $[1,a)$ is a set representing the cosets $xP$. We can define $f:\mathbb{R}^+\to[1,a)$ by setting $f(x)=xa^n$ for the unique $n\in\mathbb{Z}$ such that $xa^n\in[1,a)$. Then since for every $x\in P$ the representative can be chosen to be $1$, we have $\text{ker}f= P$. I cannot prove that $f$ is a homomorphism. Is it true that $f$ is one? It seems, that this is the obvious choice for $f$.

Answer 1

Yes, you can find a group homomorphism $f$ defined on $(\Bbb R_{>0},\cdot)$ with $\ker(f)=P$. For simplicity first apply the group isomorphism $\ln:(\Bbb R_{>0},\cdot)\to(\Bbb R,+)$, then the group homomorphism $g:(\Bbb R,+)\to(\Bbb C,\cdot)$ with kernel generated by $\ln(a)$, namely $g:x\mapsto\exp(\frac{2\pi\mathbf i}{\ln(a)}x)$; then $f=g\circ\ln:x\mapsto g(\ln x)=\exp(2\pi\mathbf{i}\log_a(x))$ is a group homomorphism $(\Bbb R_{>0},\cdot)\to(\Bbb C,\cdot)$. Its kernel is $P$ and its image is the unit circle, which indeed models the quotient group you are interested in.

Answer 2

Consider a concrete example. Let $a = 2$ and $x = 3$. Then $xP = \{ 3 \cdot 2^n \colon n \in \mathbb Z \}$. How does this contain all positive real numbers?

The homomorphism you're looking for is the one that sends elements of $\mathbb R^+$ to their cosets in $\mathbb R^+/P$.

As mentioned, make sure you're clear about the fact that the quotient group does not contain elements of the original group; it contains cosets.

Answer 3

A value of $a$ that would make this really easy to visualize is $a=10$. For instance

$$[1.2] =\{… 1200, 120, 12, 1.2, 0.12, 0.012,…\}$$

i.e. just forget the decimal point, and forget any trailing zeros.

And if you're looking for a nice, canonical representative of each coset, you can pick the member of the coset that has exactly one digit in front of the decimal point. This is basically jjagmath's suggestion that you pick the value with $1 \le x \lt 10$.

(I'll add that back when people did multiplication with slide rules or logarithm tables, people used this equivalence all the time. Those tools only provided you with the string of digits (aka mantissa), and then you had to put the decimal point in the right place separately.)