Closure of a certain subset in a compact topological group 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 $\bar{A}$.
 A: 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 $G$. 
A: 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 $g^{-1} \in \overline{A}$.
A: 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?
A: 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 $x^{-1}.g^{n_{k}} \rightarrow 1.!$ 
A: I really like Fischer's solution. I would like to post an answer based on Fischer's answer in the "filling the blank" spirit - certainly helpful for beginners like me.
We need to get help from the two theorems below (refer to section 1.15 in Bredon's Topology and Geometry):


*

*In a topological group $G$ with unity element $1$, the symmetric neighborhoods of $e$ form a neighborhood basis at $1$.

*If $G$ is a topological group and $U$ is any neighborhood of $1$ and $n$ is any positive integer, then there exists a symmetric neighborhood $V$ of $1$ such that $V^n\subset N$.


Since $A.A\subset A$, by the same argument using the continuity of the multiplication map to prove the closure of a subgroup is a subgroup, we have $\bar{A}.\bar{A}\subset\bar{A}$. Therefore, if $g^{-1}\in\bar{A}$, then $g^{-n}\in\bar{A}$ for any $n\in\textbf{N}$, and so the group $\langle g\rangle\subset\bar{A}$. Taking bar of $A\subset\langle g\rangle\subset\bar{A}$ yields $\bar{A}=\bar{\langle g\rangle}$, which means $\bar{A}$ is a subgroup due to being the closure of a subgroup.
We now show that $g^{-1}\in \bar{A}$. The case $g$ has finite order is trivial, so let $g$ has infinite order. Suppose $g^{-1}\notin\bar{A}$. Then by definition, there is a neighborhood $U'$ of $1$ that, by the homeomorphic translation, $g^{-1}U'\cap A=\emptyset$. Since the symmetric neighborhoods form a neighborhood basis of $1$, we can choose a symmetric neighborhood $U$ of $1$ such that $g^{-1}U\cap A=\emptyset$.
Since $A$ is an infinite set in a compact space, $A$ has a limit point $p$. We can choose a symmetric neighborhood $V$ of $1$ such that $V^2\subset U$. Then, $pV$, being a neighborhood of a limit point of $A$, must contain at least two distinct points $g^m$, $g^n$ ($m, n\in \textbf{N}$), i.e. $g^m=pa$, $g^n=pb$ for some $a,b\in V$. Due to $V$ being symmetric and $V^2\subset U$, $b^{-1}a=u$ for some $u\in U$, and we have $g^m=pa=pb.u\in g^n U\cap A$
Now, if $m>n$, then $g^{m-n-1}\in g^{-1}U\cap A$ (contradict with our definition of $U$). If $n>m$, then since $U^{-1}=U$, $g^m\in g^n U^{-1}$, and so $g^n\in g^m U\cap A$ (also contradict with our definition of $U$). Therefore $g^{-1}\in \bar{A}$.
