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Let $G$ be a profinite group and $ \bar{\mathbb{Z}} = \varprojlim _{n \in \mathbb{N}} \mathbb{Z}/n\mathbb{Z} \subset \prod _{n \in \mathbb{N}} \mathbb{Z}/n\mathbb{Z}$. Let $g \in G$ and define $a := (a_n + n\mathbb{Z})_{n \in \mathbb{N}} \in \bar{\mathbb{Z}}$. There is given a map $\phi:\bar{\mathbb{Z}} \to G, a \to g^a$ with $g^a := lim _{n \in \mathbb{N}} g^{a_n}$ in G and I have to show hat it is a homomorphism of groups.

My question is how to interpret/understand this limit $g^a = lim _{n \in \mathbb{N}} g^{a_n}$ ?

user267839
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  • Notice that the inverse system under which the profinite completion $\hat{\mathbb{Z}}$ is defined arises from the direct poset $D=(\mathbb{N},\leq)$ where the partial ordering $\leq$ is exactly the divisibility relation, i.e., $m\leq n$ iff $m\mid n$. Then $(g^{a_n}:n\in D)$ is a net on $G$ and hence we can discuss its convergence. Moreover, if $(G_i:i\in I)$ is an inverse system of discrete finite groups that gives rise to $G$, then $g^a$ can be explicitly written as $$(g_i:i\in I)^a = (g_i^{a_{|G_i|}}:i\in I)$$ where $|G_i|$ is the order of $G_i$. – Sangchul Lee Jul 28 '17 at 20:36
  • Ok, but in this case $G$ can be a arbitrary abstract group, can't it? Because in general the index sets $\mathbb{N}$ (of $\bar{\mathbb{Z}}$) and $I$ (of G as inverse limit $G = \varprojlim _{i \in I} G_i$) doesn't correlate in some way? – user267839 Jul 28 '17 at 21:13
  • The above explanation does not assume any relation between $I$ and $\mathbb{N}$. – Sangchul Lee Jul 28 '17 at 21:36

1 Answers1

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Assume that

$$G = \lim_{\substack{\longleftarrow\\i \in I}} G_i$$

for some inverse system $(G_i)_{i\in I}$. Then with $g = (g_i:i\in I)$,

  1. $\lim_{n\in\mathbb{N}}g^{a_n}$ converges exactly when $\lim_{n\in\mathbb{N}} g_i^{a_n}$ converges for each $i\in I$. To this end, we need to show that there exists $N$ for which $g_i^{a_n} = g_i^{a_m}$ whenever $N \mid m, n$. Notice that we can take $N = \operatorname{ord}(g_i)$ since we have $a_n \equiv a_N \equiv a_m \pmod{N}$ in this case.

  2. If $a = (a'_n+n\mathbb{Z}:n\in\mathbb{N})$ is another choice of representatives, then whenever $n$ divisible by $N = \operatorname{ord}(g_i)$ we have $a'_n \equiv a_n \pmod{N}$ and hence $g_i^{a_n} = g_i^{a'_n}$. This tells that the common value found in the previous step is independent of the choice of representatives.

This tells that $g^a = \lim_{n\in\mathbb{N}} g_i^{a_n}$ is well-defined. Showing that this map $a \mapsto g^a$ is homomorphism is now a routine work.

Perhaps a cleaner way of explaining this would be as follows:

Fix an element $g$ in $G = {\displaystyle\lim_{\longleftarrow}} G_i$ and define the map $\mathbb{Z}/n\mathbb{Z} \to G_i$ by $k+n\mathbb{Z} \mapsto g_i^k$ whenever $n$ is divisible by $\operatorname{ord}(g_i)$ (so that this map is well-defined). Then the following diagram commutes whenever all arrows make sense

$$\begin{array}{ccccc} \hat{\mathbb{Z}} & \rightarrow & \mathbb{Z}/n\mathbb{Z} & \rightarrow & G_j \\ & \searrow & \downarrow & \searrow & \downarrow \\ & & \mathbb{Z}/m\mathbb{Z} & \rightarrow & G_i \end{array}$$

where $\hat{\mathbb{Z}} \to \mathbb{Z}/n\mathbb{Z}$ are projections and vertical arrows are bonding maps.

Now let $\phi_i : \hat{\mathbb{Z}} \to G_i$ be the composition of $\hat{\mathbb{Z}} \to \mathbb{Z}/n\mathbb{Z} \to G_i$, which is well-defined and does not depend on the choice of $n$ by the above diagram. Then it follows that

$$\begin{array}{ccccc} \hat{\mathbb{Z}} & \overset{\phi_j}{\rightarrow} & G_j \\ & \smash{\underset{\phi_i}{\searrow}} & \downarrow \\ & & G_i \end{array}$$

commutes and hence (by the universal property of the inverse limit) there exists a unique map $\phi : \hat{\mathbb{Z}} \to G$ that makes the following diagram

$$\begin{array}{ccccc} \hat{\mathbb{Z}} & \overset{\phi}{\rightarrow} & G \\ & \smash{\underset{\phi_i}{\searrow}} & \downarrow \\ & & G_i \end{array}$$

commute. This $\phi$ coincides the map $a \mapsto g^a$ constructed by the limit.

Sangchul Lee
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