# If $A^p$ is ergodic, does that make $A$ ergodic? And possible circular proof?

I am reading Walters' Introduction to Ergodic Theory and am struggling to make sense of the proof of Theorem 1.10. I have an issue with understanding the overall structure of the proof, and with a component of the proof itself. Issue regarding the proof:

The sentence

So if $$a_n \neq 0$$... $$\quad$$ ... and so $$f$$ is constant a.e.

in my mind just doesn't follow. Doesn't this show that $$A^p$$ is ergodic, not $$A$$?

Issue regarding the structure:

At the beginning of the proof, an assumption is made and is later invoked to make a conclusion. How does saying

Suppose that whenever $$\gamma A^n = \gamma$$ for some $$n \geq 1$$ we have $$\gamma \equiv 1$$.

allow us to use that later on to prove that having the trivial character is the only $$\gamma \in \hat{G}$$ that satisfies $$\gamma \circ A^n = \gamma$$. It seems rather circular?

Appreciate any help.

• If $A^p$ is ergodic, then $A$ must be ergodic: the contrapositive of this statement says that if $A$ is not ergodic, then $A^p$ is not ergodic, which is easy to verify - every $A$-invariant function is also $A^p$ invariant. Regarding the second issue: the assumption that "whenever $\gamma A^n=\gamma$ for some $n\geq 1$ we have $\gamma =1$" is the second part of the "if and only if" assertion in the theorem, so it must be assumed to prove one direction of the "if and only if". – John Griesmer Jan 13 at 3:07
• John, that clears everything up! Thank you so much. Do you want to write up the answer so I can mark it as correct? – Alex Hiller Jan 13 at 6:10

If $$A^p$$ is ergodic, then $$A$$ must be ergodic: the contrapositive of this statement says that if $$A$$ is not ergodic, then $$A^p$$ is not ergodic, which is easy to verify - every $$A$$-invariant function is also $$A^p$$ invariant.
Regarding the second issue: the assumption that "whenever $$\gamma A^n=\gamma$$ for some $$n\geq 1$$ we have $$\gamma = 1$$" is the second part of the "if and only if" assertion in the theorem, so it must be assumed to prove one direction of the "if and only if".