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Let $\widehat{F_2}$ be the profinite free group of rank 2, and let $\widehat{\mathbb{Z}}$ be the profinite completion of $\mathbb{Z}$, and $\widehat{\mathbb{Z}}^\times$ its group of units.

For $n\in\widehat{\mathbb{Z}}^\times$, and $x\in\widehat{F_2}$, what does it mean to say $x^n$? (does this even make sense? If it doesn't, does it make sense "up to conjugacy"?)

Also, is there a name for elements of $\widehat{\mathbb{Z}}$? (elements of $\mathbb{Z}_p$ are called $p$-adic integers)

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Elements of $\widehat{\mathbb{Z}}$ are called profinite integers. The profinite integers have a universal property in the category of profinite groups in exactly the same way that the integers have a universal property in the category of groups: namely, $\widehat{\mathbb{Z}}$ is the free profinite group on one generator. This means precisely that elements $g \in G$ of a profinite group $G$ are the same thing as (continuous) homomorphisms

$$\widehat{\mathbb{Z}} \ni n \mapsto g^n \in G.$$

These homomorphisms can be defined in various equivalent ways: for example, you know what $g^n$ means for ordinary integers $n$, and you can use continuity to figure out what it must mean for profinite $n$ from here.

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  • $\begingroup$ So would I be correct in saying the following: Given an element $g\in G$ and $n\in\widehat{\mathbb{Z}}$, $n$ is a compatible tuple of all its images $n_m$ in $\mathbb{Z}/m\mathbb{Z}$ for all $m \ge 1$. Then suppose $G$ is the inverse limit of finite groups $G_i$, so $g = (g_{G_i})_i$, then $g^n$ is essentially just $(g_{G_i}^{n_{ord(g_{G_i})}})_i$ $\endgroup$ – oxeimon Aug 31 '15 at 18:03
  • $\begingroup$ @oxeimon: yes, that's correct. $\endgroup$ – Qiaochu Yuan Aug 31 '15 at 20:21

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