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What do subgroups of the Prufer 2-group $\mathbb{Z}_2(2^{\infty})$ look like?

Let's think of $\mathbb{Z}_2(2^{\infty})$ as the dyadic rationals modulo $1$

EDIT: An example of a subgroup and its subgroup would be helpful.

While we're at it; is there a better way of writing the dyadic rationals modulo $1$?

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    $\begingroup$ You could write a group presentation: $\langle g_n\mid g_{n+1}^2=g_n,g_1^2=e\rangle$, or describe it as a subset of the unit circle in the complex plane: $\{e^{2\pi m/2^n}\mid m,n\in \Bbb N\}$. But I don't think the group itself has a generally agreed-upon symbol. $\endgroup$ – Arthur Jan 31 '18 at 14:48
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    $\begingroup$ The simple answer is that all proper subgroups are finite and cyclic, and there is one such of order each power of $2$. $\endgroup$ – Derek Holt Jan 31 '18 at 15:05
  • $\begingroup$ Thanks @Arthur I'm looking to avoid the unit circle representation at the moment and find a notation that makes it clear I'm talking about dyadic rationals modulo $1$. $\endgroup$ – samerivertwice Jan 31 '18 at 15:19
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I don't know whether it is better, or not, but you could write the dyadic rationals are the direct limit of infinite cyclic subgroups of the rational numbers,i.e., $$ \varinjlim \left\{2^{-i}\mathbb{Z}\mid i = 0, 1, 2, \dots \right\} $$ Concerning subgroups of Prüfer $p$-groups, have a look at this post:

Characterising subgroups of Prüfer $p$-groups.

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  • $\begingroup$ Does the modulo 1 relation get involved in its this dyadic rationals representation? $\endgroup$ – samerivertwice Feb 1 '18 at 7:54

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