# What do subgroups of $\mathbb{Z}_2(2^{\infty})$ look like?

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 $1$?

• 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. – Arthur Jan 31 '18 at 14:48
• The simple answer is that all proper subgroups are finite and cyclic, and there is one such of order each power of $2$. – Derek Holt Jan 31 '18 at 15:05
• 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 $1$. – samerivertwice Jan 31 '18 at 15:19

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.