Prove that if $0$ and $M$ are the only submodules of $M$ stabilized by every endomorphism of $M$ then $M$ is semisimple and homogeneous.
I am thinking of using 'If every submodule of $M$ is direct summand then $M$ is semisimple.' So I start with a submodule $M'$. Now using that there exists an endomorphism which does not stabilize $M'$ I should be able to produce $M''$ such that $M =M' \oplus M''$. I am stuck here.
On
As Andrew Hubery’s nice example shows, this is incorrect in general, but I will sketch out the case for proving it when the socle $\mathrm{soc}(M)\neq\{0\}$.
Firstly, you should find not easy to prove the socle is invariant under all endomorphisms (since the homomorphic image of a semisimple module is semisimple.)
By the same token, you can see that a homogeneous component is invariant under endomorphisms, because their image can only contain copies of the simple module from the domain of the homomorphism.
Under the assumption that only two submodules are invariant, you can see that there is only one possibility if the socle is nonzero.
This is false in general. Consider the rationals $\mathbb Q$ as a $\mathbb Z$-module. Multiplication by any rational $q$ is an endomorphism, and so the only submodules which are stable under all endomorphisms are $0$ and $\mathbb Q$. However, $\mathbb Q$ is not semisimple; it does not even have any simple submodules or simple quotient modules.