Generalized Clifford's Theorem A typical statement of Clifford's theorem is the following:
Let V be a finite dimensional irreducible representation of a group G, and let N be a normal subgroup of finite index in G. Then the restriction of V to N is a semisimple.
The question I am trying to answer is the following:
Let V be an irreducible representation of a group G and let N be a normal subgroup of G. Show that the restriction of V to N is semisimple.
There are no other restrictions on G,V or N. Is there any way to get this result assuming Clifford's Theorem? Or is there a way to prove this in a similar manner?
 A: The sum of (even an infinite set of) irreducible submodules is semisimple, and the sum of all irreducible $N$-submodules of $V$ is a $G$-submodule, which must be $0$ or $V$, since $V$ is irreducible as a $G$-module. So if the restriction of $V$ has any irreducible submodules at all then it must be semisimple. 
If $V$ is finite dimensional then this is the case, even if $N$ has infinite index.
If $N$ has finite index in $G$ then it's also the case. If $v$ generates $V$ as a $G$-module, then $\{vg_1,\dots,vg_n\}$ generates as an $N$-module, where $\{g_1,\dots,g_n\}$ is a set of coset representatives. In particular, the restriction of $V$ is finitely generated, and so has a maximal submodule $U$. But then $Ug_i$ is also a maximal submodule, and the natural map $V\to\bigoplus V/Ug_i$ is a nonzero map to a semisimple $N$-module, whose kernel is a $G$-submodule and therefore zero, since $V$ is irreducible. But every submodule of a semisimple module is semisimple. 
But there are infinite dimensional examples for normal subgroups of infinite index where the restriction has no irreducible submodules (and so is not semisimple). For example, let $k$ be any field, let $V=k(t)$, the field of rational functions, let $G$ be the multiplicative group of $k(t)$ acting on $V$ by multiplication, and let $N$ be the cyclic subgroup generated by $t$. Then $V$ is an irreducible $kG$-module, but has no irreducible $kN$-submodule, since the $kN$-submodule generated by any $f(t)\in k(t)$ has a nontrivial proper submodule generated by $(t+1)f(t)$.
