Let $G$ be a metrizable compact topological group. Suppose for some $g_0 \in G$ the translation $L_{g_0}$ is topologically transitive. Prove that $G$ is abelian.
Where $ L_{g_0}g = g_0g$.
I think I found a solution to this problem, but I can't see whether I am using the compactness of the group, maybe someone can help. This is my proof:
We will show that given a metric $d$ on the group, $\forall \epsilon>0,\; d(\, g_1g_2\, ,\, g_2g_1\,)<\epsilon$.
If a topological group $G$ admits a topologically transitive translation $L_{g_0}$ then it's minimal, i.e. the orbit is dense $\forall g \in G$. Consider the orbit $\{L^n_{g_0} id\}_{n\in \mathbb{Z}}$, where $id$ is the identity in the group. By continuity of the operation we find:
$$\exists n_1' : d(L_{g_0}^{n_1'}\, id \,,\, g_1 ) < \delta_1' \implies d(g_2L_{g_0}^{n_1'}\, id \,,\, g_2g_1 ) < \frac{\epsilon}{4}$$ $$\exists n_1'' : d(L_{g_0}^{n_1''}\, id \,,\, g_1 ) < \delta_1'' \implies d(L_{g_0}^{n_1''}\, id \,g_2 \,,\, g_1g_2 ) < \frac{\epsilon}{4}$$
Let's call $n_2$ the number between $n_2'$ and $n_2''$ that would give a distance less than $\delta_1 = \min\{\delta_1',\delta_1''\}$.
Similarly
$$ \exists n_2': d(L_{g_0}^{n_2' + n_1} id \,,\, g_2L_{g_0}^{n_1} id) < \frac{\epsilon}{4}$$ $$ \exists n_2'': d(L_{g_0}^{n_1 + n_2''} id\,,\, L_{g_0}^{n_1}id \, g_2) < \frac{\epsilon}{4} $$ Where $L_{g_0}^{m+n} id = L_{g_0}^{n+m}id = L_{g_0}^n id \; L_{g_0}^m id $.
We define $n_2$ similarly to $n_1$. Finally: $$d(\, g_1g_2\, ,\, g_2g_1\,)\leq d(\, g_1g_2, L_{g_0}^{n_1+n_2}id\,) + d(\, L_{g_0}^{n_2+n_1}id\,, g_2g_1\,)\leq$$ $$ \leq d(\,g_1g_2,L_{g_0}^{n_1}id\,g_2\,) + d(\, L_{g_0}^{n_1}id\, g_2,L_{g_0}^{n_1 + n_2} id\,) + d(\,L_{g_0}^{n_2 + n_1} id,g_2 L_{g_0}^{n_1}\,) + d(\,g_2 L_{g_0}^{n_1},g_2g_1 \,) < 4\,\frac{\epsilon}{4} = \epsilon$$ Which concludes the proof.
Sorry if the formatting isn't perfect but i am a first-time poster. Anyway thanks for the help.