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Suppose that $G$ is a compact Hausdorff topological group and that $g\in G$. Consider the set $A=\{g^n : n=0,1,2,\ldots\}$ and let $\bar{A}$ denote the closure of $A$ in $G$.

Is it true that $\mathbf{\bar{A}}$ is a subgroup of $\mathbf{G}$?

From continuity of multiplication and the fact that $A\cdot A\subseteq A$ it is clear that $\bar{A}\cdot\bar{A}\subseteq\bar{A}$. therefore, $a,b\in \bar{A}$ yields $a\cdot b\in \bar{A}$. However, I am having trouble showing that inverses of elements in $\bar{A}$ are also in $\bar{A}$.

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It is sufficient to show that $g^{-1} \in \overline{A}$, then $\langle g\rangle \subset \overline{A}$, and $\overline{A}$ is the closure of a subgroup, hence a subgroup.

If $g$ has finite order, it is trivial that $g^{-1} \in A = \overline{A}$, so let's suppose that $g^n \neq 1$ for $n \neq 0$.

If $g^{-1} \notin \overline{A}$, then there is a symmetric open neighbourhood $U$ of $1$ with $g^{-1}U \cap A = \varnothing$.

Then, for all $n \in \mathbb{N}$, we have $g^nU \cap A = \{g^n\}$. Otherwise, if $g^k \in g^nU$ for $k \neq n$, by symmetry of $U$ we can assume that $k > n$, then $g^{k-n-1} \in g^{-1}U\cap A$, which contradicts the choice of $U$.

Now, $A$ is an infinite set, hence has an accumulation point $p$, since $G$ is compact. Let $V$ a symmetric open neighbourhood of $1$ with $V\cdot V \subset U$. Since $p$ is an accumulation point of $A$, $pV$ contains at least two points $g^k$ and $g^n$, $n \neq k$ of $A$. But then $g^k \in g^nU \cap A$. That contradicts the above, hence $g^{-1} \in \overline{A}$.

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up vote 3 down vote accepted

Let $B=\{g^{n}:n\in\mathbb{Z}\}$. Clearly $\bar{B}$ is also a subgroup of $G$.

If $1$ is an isolated point in $\bar{B}$ then all points of $\bar{B}$ are isolated, which means that $\bar{B}$ is compact and discrete, and hence finite. Thus, $g^{n}=1$ for some $n$ and so $\bar{A}$ is a subgroup of $G$.

On the other hand, if $1$ is a limit point in $\bar{B}$ then for any symmetric neighborhood $V$ of $1$ there is a positive integer $n$ such that $g^{n}\in V$ ($n>0$ is possible since $V$ is symmetric). Then $g^{n-1}\in(g^{-1}V)\cap A$, and since the $g^{-1}V$ form a neighborhood basis at $g^{-1}$ we have that $g^{-1}\in \bar{A}$. This means that $\bar{A}=\bar{B}$ and so $\bar{A}$ is a subgroup of $G$.

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Very nice. ${}$ – Mariano Suárez-Alvarez Jul 26 '13 at 0:29

Let me suppose your group is second-countable, so that we need only worry abour sequences.

It is enough to show that for every $g\in G$ we have that $g^{-1}$ is an element of the closure of $\{g^i:i\geq0\}$.

So pick any convergent subsequence of $(g^i)_{i\geq0}$, say $(g^{n_i})_{i\geq0}$, and let $h\in G$ be its limit. Then $(g^{n_i-n_{i-1}-1})_{i\geq1}$ is another subsequence. What is its limit?

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The general case follows similarly. – Mariano Suárez-Alvarez Jul 25 '13 at 22:03
Thanks, +1 to your answer – John Jul 26 '13 at 1:16

I am editing what I said before which mistakenly assumed $A$ above is a subgroup.

Here is the answer. Assume that $g$ is not nilpotent (in that case there is nothing to show).

Claim: The point {1} is not isolated in $\bar{A}$.

The claim implies that there is a non-trivial sequence in $\bar{A}$ hence in $A$ which converges to 1. Then writing this sequence as $\{ g^{k_{i}} \}$ we get $\lbrace { g^{k_{i}-1} \rbrace} \rightarrow g^{-1}$. Hence $g^{-1} \in \bar{A}$ and thus $\bar{A}$ is the closure of the subgroup generated by $A$ and hence is a subgroup (which is a simple fact).

To prove the claim. The set $A$ has atleast one limit point (because of compactness). Call it $x$. Then for any sequence $\{ g^{n_{k}} \} \rightarrow x$. Hence $x^{-1}.g^{n_{k}} \rightarrow 1.!$

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How does this help answer the question? – Chris Eagle Jul 25 '13 at 21:51
If $x=1$ in the additive group $\mathbb R$... – Mariano Suárez-Alvarez Jul 25 '13 at 21:51
@ChrisEagle : I apologize, I thought A is a subgroup but I now notice that it is not. It is a semi-group. – DBS Jul 25 '13 at 21:53

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