Ytukyg

This question has already been asked here: Group representation is semisimple iff restriction to subgroup of finite index is semisimple, but only one direction of the proof is provided. I'm asking here about the other one. Also, this question, Semisimplicity of restriction: Representation theory, asks the same as mine, but I don't understand the comment by Turion: if the restriction will be the direct sum of $2$ $1$-dimensional representations, won't it also be semisimple, as a direct sum of $2$ irreducible representations? Namely:

If $G$ is a group, $k$ a field, $E$ a $k$-vector space, $rho: G to GL(E)$ a semisimple linear representation, $H triangleleft G$, $[G : H] < +infty$. Prove (or disprove!) that $pi := mathrm{Res}_{H}^{G}(rho)$ given by $pi(h)v = rho(h)v$ is also semisimple.

The hint for the exercise in the book An Introduction to the Representation Theory of Groups by Emmanuel Kowalski gives the following hint: One can assume that $rho$ is irreducible - show that there exists a maximal semisimple subrepresentation of $pi$.

So as soon as I saw "maximal", I thought about Zorn's lemma. The problem is, I have no idea how to use the finiteness of the index or the irreducibility of $rho$. Here's what I came up with: let $mathcal{M}$ denote the set of all semisimple subrepresentations of $pi$. $0 in mathcal{M}$ so $mathcal{M}$ is non-empty. Then, let $mathcal{L} = lbrace sigma_{alpha}: H to GL(E_{alpha}) rbrace_{alpha in A}$ be a chain in $mathcal{M}$. Then obviously $sigma: H to GL(bigcup_{alpha in A}E_{alpha})$ is a subrepresentation of $E$. So now my idea was to show that $sigma$ is completely reducible.

Let $F_{1} leq bigcup_{alpha in A}E_{alpha}$ be a subrepresentation. Then $F_{1} cap E_{alpha}$ is also a subrepresentation of $E_{alpha}$, so by semisimplicity, i.e. complete reducibility of $E_{alpha}$ there exists a subrepresentation $F_{alpha} leq E_{alpha}$ such that $F_{1} cap E_{alpha} oplus F_{alpha} = E_{alpha}.$ I'd like to take $F_{2} := bigcup_{alpha in A} F_{alpha}$ and show $bigcup_{alpha in A} E_{alpha} = F_{1} oplus F_{2}$.

The problem: $lbrace F_{alpha} rbrace_{alpha in A}$ is not a chain, so I don't even know that $F_{2}$ would be a vector subspace. I could take $F_{2} = sum_{alpha in A} F_{alpha}$, but I wouldn't know that $F_{1} cap F_{2} = 0$.

Another problem: I have no idea how to use the fact that $H$ has finite index in $G$ or that $rho$ is irreducible, which probably tells me I'm not on the right track.

Can anyone give me an explanation of what a proof of this statement would look like, or in case the statement is false, an explanation of why it's false?