Convergent subsequence of $x^n$ to the identity element in a compact group

Question

The following question was recently asked by the user @Héhéhé and was downvoted twice (as usual, without any comments), which led the OP to censor himself by deleting his question. As far as I know, this question is interesting and nontrivial. I will provide an answer within a few hours if nobody answers, but I am confident that the downvoters will propose their own answer in the meantime.

Consider the following statement:

Let $G$ be a compact group with identity element $e$. For all $x \in G$, the sequence $(x^n)_{n\in \mathbb N}$ has a subsequence that converges to $e$.

Is it true? If yes can you provide a proof (or a source to a proof)? If no can you provide a counter-example?

If necessary, we can restrict ourselves to the case where $G$ is a metric space.

Answer 1

Consider the closure $H$ of $L=\{x^n, n\in \mathbb{Z}\}$, it is a closed (compact) subgroup of $G$. If the connected component of the identity $H_0$ of $H$ is the identity, then $H$ is finite and the order of $x$ is finite. If $H_0$ is not the identity, then $L\cap H_0$ is dense in $H_0$, there exists $x^{n_p}$ which converges towards $e$. By extracting a subsequence from $x^{n_p}$ or $x^{-n_p}$ we can construct a subsequence of $\{x^n,n\in\mathbb{N}\}$ which converges towards $e$.

Answer 2

For the case that $G$ is a metric space the following should work:

Fix $x \in G$. As $G$ is compact there is a sequence $n_1< n_2 < \dots$ in $\mathbb{N}$ such that $(x^{n_k})$ is convergent to $y$, say. As inversion is continuous $x^{-n_k} \to y^{-1}$ $(k \to \infty)$. To each $j \in \mathbb{N}$ there is some $n_{k_j}$ such that $m_j:=n_{k_j}-n_j > j$. Now $m_j \to \infty$ and $x^{m_j} \to y y^{-1}=e$ as $j \to \infty$.