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Let $G$ be a compact topological group with normalized Haar measure $\mu$. An element $g\in G$ is an ergodic element if the mapping $L_g:G \to G $ with $x\mapsto gx$ is an ergodic map. The set of ergodic elements of a group $G$ is denoted by $E(G)$

Is there a compact topological group or a compact Lie group for which the set of all ergodic elements have intermediate measure namely we have $0<\mu(E(G)<1$?

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    $\begingroup$ Yes, the finite group $\mathbf{Z}/4\mathbf{Z}$ is an example. The set of ergodic elements has cardinal 2, hence measure $1/2$. $\endgroup$
    – YCor
    Commented Sep 20, 2023 at 11:22
  • $\begingroup$ @YCor Yes thank you that is perfect. I was inspired by the circle case. The ergodic element has full measure. So what about we add the extra assumption connected lie group? BTW is there an example of empty or zero measure set of ergodic elements? $\endgroup$
    – Ali Taghavi
    Commented Sep 20, 2023 at 11:27
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    $\begingroup$ You mean "in the connected case". Every finite group is a Lie group. $\endgroup$
    – YCor
    Commented Sep 20, 2023 at 11:45
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    $\begingroup$ For a connected compact nonabelian Lie group there are no ergodic elements. For a connected compact abelian Lie group almost all elements are ergodic. $\endgroup$
    – Moishe Kohan
    Commented Sep 20, 2023 at 19:00
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    $\begingroup$ My previous comment answers the modified question: there are no such groups. $\endgroup$
    – Moishe Kohan
    Commented Sep 21, 2023 at 4:34

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