Why the semisimplicity of representation is equivalent to the semisimplicity of the element?

Consider $$1\in \mathbb Z \subset \widehat {\mathbb Z}$$(it's the topological generator of $$\widehat {\mathbb Z}$$). Let $$l$$ be a prime. Suppose there is a continuous representation $$\rho:\widehat {\mathbb Z}\to \operatorname{Aut}_{\mathbb Q_l}(V)$$ for some finite-dimensional vector space $$V$$ over $$\mathbb Q_l$$.

Denote $$u=\rho(1)$$, then I need to prove:

$$\rho$$ is semisimple $$\Leftrightarrowu$$ is semisimple.

I know the semisimplicity of $$\rho$$ is equivalent to: $$V$$ is a direct sum of irreducible representations, and the semisimplicity of $$u$$ is that $$u$$ is diagonalizable.

But how can we connect these two properties? Could you give some hints for me? Thanks.

• "semisimplicity of $u$ is that $u$ is diagonalizable". Are you sure that's the definition? (It makes a proof easier, but:) Since $\mathbb Q_l$ is not algebraically closed, a general vector space $V$ over it will have endomorphisms $f$ which are semisimple (which for me means, $V$ is semisimple as $\mathbb Q_l[f]$-module) without $f$ being diagonalisable (over $\mathbb Q_l$). In fact, every proper field extension of $\mathbb Q_l$ will give examples: take, as $f$, multiplication with any element not in the ground field. Aug 13, 2022 at 15:38
• @TorstenSchoeneberg Thanks for pointing out! Yes, the definition should be changed as what you've said. Actually, semisimple is equivalent to "diagonalizable over the algebraic closure of $\mathbb Q_l$", rather than over $\mathbb Q_l$. Aug 14, 2022 at 14:22

1. If $$u$$ is semisimple, then every element in the image of the representation is semisimple. (Via topological generation.)
2. If two semisimple endomorphisms of a finite-dimensional vector space $$V$$ over a perfect field commute, then there is a decomposition of $$V$$ into subspaces which are irreducible simultaneously for both endomorphisms. Cf. e.g. Simultaneous semisimplicity of commuting endomorphisms, A family of commuting endomorphisms is semisimple if each element is semisimple, or Bourbaki Algebra VII §5 no. 8. --- And the group we are letting act here is abelian.
• Elaborating on 1: We know $\rho(\widehat{\mathbb Z})$ is the closure of $\rho(\mathbb Z)$ [this needs a proof]. But being semisimple is a closed condition, so if $\rho(\mathbb Z)$ consists of semisimple elements, so does $\rho(\widehat{\mathbb Z})$. Aug 14, 2022 at 8:03