I'm trying to figure out what all the subgroups of $\mathbb{R}$ that are generated by two elements are, but I'm not sure how to get started?

Anybody have any hints that could point me in the right direction?

  • $\begingroup$ They are the groups of the form $\langle a,b\rangle$ with $a,b\in\Bbb{R}$. There are very many of them. What kind of answer are you looking for? $\endgroup$ – Servaes Oct 23 '15 at 22:53

Well $\newcommand{\RR}{\mathbb{R}}$$\newcommand{\ZZ}{\mathbb{Z}}$lets start with subgroups of $\RR$ generated by 1 element. These are by definition the cyclic subgroups of $\RR$. Suppose $x\in\RR$, then the subgroup of $\RR$ generated by $x$ is the set $\langle x\rangle = \{\ldots,-3x,-2x,-x,0,x,2x,3x,\ldots\}$. This is because $\RR$ is an additive group, so if you have $x$, then you have $-x$, and hence you have $x + (-x) = 0, x+x = 2x, -x + (-x) = -2x,x+x+x = 3x,\ldots$.

Okay, so the cyclic subgroups of $\RR$ are pretty easy to describe, they are in fact either trivial (if $x = 0$), or isomorphic to $\ZZ$. To see this, describe a homomorphism from $\ZZ$ to $\langle x\rangle$ given by sending $1\mapsto x$.

Now suppose $x,y\in\RR$. Describing the two-generator subgroups of $\RR$ requires somewhat more machinery.

Since $\RR$ is an abelian group, a two-generator subgroup is precisely the image of a homomorphism $f : \ZZ^2\rightarrow \RR$. Since $\RR$ is torsion free, $f(\ZZ^2)$ is torsion-free, so by the fundamental theorem of finitely generated abelian groups, it's characterized by one number - it's rank. In other words, $f(\ZZ^2)$ is either trivial, or isomorphic to $\ZZ$, or isomorphic to $\ZZ^2$. All three examples can occur! Let $x = (1,0)\in\ZZ^2$, and $y = (0,1)\in\ZZ^2$.

  1. If $f(x) = f(y) = 0$, then of course $f(\ZZ^2) = 0$.

  2. If $f(x) = 0$ but $f(y)\ne 0$, then $f(\ZZ^2) = \ZZ f(y)$ is cyclic generated by $f(y)$. Similarly if $f(x) \ne 0$ but $f(y) = 0$.

  3. If $f(x),f(y)$ both are not 0, then the image can be either rank 1 or 2. In this case it is rank 2 if and only if $f(x),f(y)$ are $\ZZ$-linearly independent. Ie, there do not exist integers $a,b\in\ZZ$ with $af(x) + bf(y) = 0$. Exercise: Prove this. Also if the rank of $f(\ZZ^2)$ is 1, explicitly find a generator for $f(\ZZ^2)$ in terms of $a$ and $b$. (Hint: Think gcd)

  • $\begingroup$ +1: One critique. $f(x),f(y)$ are linearly independent if and only if for $a,b\in\Bbb Z$ such that $af(x)+bf(y)=0,$ we have $a=b=0.$ One can then show that $f(x)$ and $f(y)$ are linearly independent if and only if we cannot write either as a rational multiple of the other. $\endgroup$ – Cameron Buie Oct 23 '15 at 23:12
  • $\begingroup$ @CameronBuie You edited my answer and now the $\RR$'s are not rendering... $\endgroup$ – oxeimon Oct 23 '15 at 23:15
  • $\begingroup$ Oh, dear! They rendered in the preview. My apologies! $\endgroup$ – Cameron Buie Oct 23 '15 at 23:18
  • $\begingroup$ @oxeimon, I am going to require some clarification. Specifically, 1) how does $\mathbb{R}$ torsion free imply that $f(\mathbb{Z}^{2})$ is torsion free? 2) What is $\mathbb{Z}f(y)$? That notation looks really weird to me. $\endgroup$ – ALannister Nov 2 '15 at 1:49
  • $\begingroup$ @JessyCat (1) Suppose $x\ne 0$ is a torsion element of $f(\mathbb{Z}^2)$. Then, $nx = 0$ in $f(\mathbb{Z}^2)$ for some $n\in\mathbb{Z}$. But that means $nx = 0$ in $\mathbb{R}$, but $\mathbb{R}$ is torsion free, so $n = 0$. Hence $f(\mathbb{Z}^2)$ is torsion-free. (2) $\mathbb{Z}f(y) = \{\ldots,-3f(y),-2f(y),-f(y),0,f(y),2f(y),3f(y),\ldots\}$ $\endgroup$ – oxeimon Nov 2 '15 at 2:48

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