I want to learn the correct notation and language with which to communicate clearly about a quotient on the multiplicative group $G=(\Bbb Q^+,\times)$

In particular, I want to know the notation with which to clearly identify the cosets and the unique indexing elements.

The equivalence relation is $x\sim y\iff\exists i,j\in\Bbb Z:2^i \times3^j\times x=y$

Let $N$ be the 3-smooth numbers then $1\sim n\iff n\in N$ so I think this is a normal subgroup of $G$ and the equivalence class of the identity element.

Then the multiplicative group generated by the primes greater than $3$ is its dual and the quotient group. So I think I would write $G/N=Q$

Then every element of $G$ has a unique representation as $q\times n:q\in Q, n\in N$

All good so far, what I'm not clear on now is how to express (in notation) the unique representatives and their cosets.

Tentatively I think I can write $q\in Q$ and $[q]=q\times N$.

Is $Q\subset G$ or are $[q]\in Q$? i.e. does $Q$ contain singletons or cosets?

Assuming $Q$ contains cosets, how do I identify a (the) set of unique indexing elements, and its elements?

I have in mind the unique set satisfying $\lvert x\rvert_2=1=\lvert x\rvert_3$

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    $\begingroup$ How is $G$ even a group to begin with? $\endgroup$ Nov 8 '18 at 11:39
  • $\begingroup$ @TobiasKildetoft you were of course totally right, v. silly of me. Fixed now. $\endgroup$ Nov 8 '18 at 11:47
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    $\begingroup$ You should better denote the positive rationals by $\;\Bbb Q^+\;,\;\;\Bbb Q_+\;$ or something like this. Trying to use less cumbersome notation can make things simpler to understand. $\endgroup$
    – DonAntonio
    Nov 8 '18 at 11:54
  • $\begingroup$ @DonAntonio ok I've done that, thanks. $\endgroup$ Nov 8 '18 at 12:17

First, it is never true formally that $\;Q=G/N\subset G\;$ . The elements of each set are just different.

Next, if you already chose $\;q\in Q=G/N\;$ , then there is no need to use $\;[q]\;$ , which is a notation usually reserved, in this context, to denote equivalence classes for an element in the original set. Thus, if you take $\;g\in G\;$ , then it'd make sense to talk about $\;[g]:=gN\in G/N=Q\;$ (no need of multiplication sign $\;g\times N\;$ , which could be confusing) .

The elements of $\;Q=G/N\;$ are cosets, which is the name given in this context to the equivalence classes in $\;G\;$ we get fro the equivalence relation determined by $\;N\lhd G\;$ .

Thus, a singleton in $\;Q\;$ could be denoted by $\;\left\{\;[g]\;\right\}=\left\{\;gN\;\right\}\;$ , for any $\;g\in G\;$ .

I think you're having some trouble understanding all this, and you chose a rather problematic example of subgroup in a rather problematic group (for a beginner). There are way simpler examples of groups and normal subgroups to work your way through these definitions.

  • $\begingroup$ Thank-you. Which are you saying is not formally true: $Q=G/N$ or $G/N\subset G$? I think the 2nd because $G/N$ is more like $\subset\mathcal P(G)$? There is one further part I should have asked which is naturally part of the same question... Every $gN$ contains a unique indexing element $x$ which satisfies $\lvert x\rvert_2=\lvert x\lvert_3=1$. What would be a natural way to talk of this set of unique indexing elements? I don't find this group hard at all because it's just $\Bbb Z^{<\omega}$. I understand it but I need to communicate what I understand to others! $\endgroup$ Nov 8 '18 at 14:11
  • $\begingroup$ (and the quotient is $Q=\Bbb Z^{<\omega}/\Bbb Z^2$)... actually my apologies, the question did ask for the "unique indexing element", but I failed to identify what I meant by that. $\endgroup$ Nov 8 '18 at 14:20
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    $\begingroup$ @RobertFrost By your definition, $\;Q=G/N\;$ . Nobody can say this isn't true: it is just a definition. What isn't true is $\;G/N\subset G\;$ formally . And no, again: $\;G/N\;$ CANNOT be an element of the power set $\;\mathcal P(G)\;$ of G, since its elements are not elements of $\;G\;$ .... Next, I really don't understand what you mean by "indexing element", and even less what $\;|x|_2\;,\;\;|x|_3\;$ can possibly mean within this context. With all due respect, I think you have big problems understanding these rather basic things of elementary group theory. $\endgroup$
    – DonAntonio
    Nov 8 '18 at 16:46
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    $\begingroup$ @RobertFrost I think what you call "indexing element" of a coset is really a representative of the equivalence class. In general, there is no canonical choice of such representatives, so there is no notation for that. $\endgroup$
    – Arnaud D.
    Nov 9 '18 at 11:41
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    $\begingroup$ @RobertFrost Yes, you are. I guess DonAntonio misread your comment as $Q\in \mathcal{P}(G)$, which, as he said, is false. $\endgroup$
    – Arnaud D.
    Nov 9 '18 at 13:34

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