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Does there exist an endomorphism of $(\mathbb Z_p,+)$ of finite order different of $x\mapsto\xi x$ where $\xi$ is a root of unity in $\mathbb Z_p$?

Thanks in advance

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1 Answer 1

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I claim that any endomorphism of $\mathbb{Z}_p$ is just multiplication by a constant.

Let $f$ be an endomorphism of $\mathbb{Z}_p$. First observe that $p^n f(x) = f(p^n x)$, so $f$ takes $p^n \mathbb{Z}_p$ to $p^n \mathbb{Z}_p$. This shows that $f$ is continuous. For any $x \in \mathbb{Z}_p$, we can find a sequence of integers $x_1, x_2, \ldots$ such that $x_n \to x$. By continuity, we have $x_n f(1) = f(x_n) \to f(x)$ and therefore $f(x) = x f(1)$.

If $f$ moreover has finite order, then $f(1)$ has to be a root of unity.

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