Question: If $(X,E,f,\mu)$ is ergodic ppt, is $(X,E,f_k,μ)$ also ergodic for any $k \in \Bbb N$?
I believe this to be true; however, I am having difficulty in proving it.
When trying to prove ergodicity for $f_k$, I was trying to show that $\mu(f^{-k} E \Delta E ) = 0$ implies $\mu(E)=0$ but this didn't seem to work.
The answer is no.
Take $X=\{0,1\}$ and $f(0)=1$, $f(1)=0$. This is clearly ergodic. However $f^2$ is the identity map which is not.
More generally: If you aware of the structure of probability measure preserving systems, then you may know that any system $(X,T)$ admits a "Kronecker factor" which we denote by $(Z,S)$ (This factor corresponds to the minimal $\sigma$-algebra on which all the eigenvectors of $T$ are measurable). If $X$ is ergodic, $Z$ can be given the structure of a group and $S$ is a multiplication by an element in the group. (In my example $X=Z$ is the group $\mathbb{Z}/2\mathbb{Z}$, and $f$ is a translation by $1$).
One can prove (not so trivial) that $(X,T^n)$ is ergodic for all $n$ implies that $Z$ is a connected group.
Let me say one last thing. If $(X,f)$ is a system such that $(X,f^n)$ is ergodic for all $n$ then $(X,f)$ is called totally ergodic.