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The HW problem: If $T^m$ is ergodic, show $T^{m^2}$ is ergodic. (Where we can assume $T$ is measure-preserving transformation on a probability space, I think. It wasn't in the problem, but everything we've done has had that hypothesis).

So (and this may sound stupid), but I believe the problem is true. I began by thinking of ergodic transformations whose higher power weren't ergodic, and eventually arrived at $n$-cycles in $S_n$. I realized that the theorem implied the fact that if an $n$-cycle raised to the $m$ was still an $n$-cycle, then it would still be an $n$-cycle if raised to the $m^2$. However other than this, I've yet to make much progress. After all, this result has nothing to do with $m^2$. We could have chosen $m^3$, for example.

I guess what I'm asking is for a little inspiration. Definitely not the answer. I just kinda want a push in the right direction. I see reason as to why it could be true, but no concrete way as to how I should be approaching it.

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Rikimaru
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1 Answers1

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Alright, I solved this already, I think. So, if there is an invariant set $B$ for $T^{m^2}$, then we can think of $T^{-1}$ acting on $\{B,T^{-1}B, ..., T^{m^2 - 1}B\}$ by permuting the elements. That is to say, we can view $T^{-1}$ as the $m^2$-cycle in $S_{m^2}$ that looks like $(0123...m^2 - 1)$. So, we reduce the problem to the special case I had considered, and hence we're done since the special case can be proved by just playing with the cycles.

Rikimaru
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