If a left module M is co-isosimple and semi-Hopfian, then M is simple. I have tried to prove following theorem;
The followings are equivalent for a left module M:

*

*M is simple,

*M is co-isosimple and co-regular,

*M is co-isosimple and semi-Hopfian,

*M is co-isosimple and discrete.


So far, I have proved (1$\Rightarrow$4$\Rightarrow$2$\Rightarrow$3), but I couldn't prove (3$\Rightarrow$1) which is the last step.

Definition 1: A non-zero module M is co-isosimple if it is isomorphic to all its non-zero quotients.

Definition 2: A left module M is a semi-Hopfian module if for every epimorphism p:M$\longrightarrow$M we have that ker(p) is a direct summand of M.

In addition, I've seen a theorem in another article that, a left module M is semi-Hopfian if and only if for any submodule N of M which satisfies M/N is isomorphic to M is a direct summand of M.

By using just these information, I can say that since M is co-isosimple, then for every submodule N of M, M/N is isomorphic to M, and by the information that I wrote after definitions, every submodule of M is a direct summand of M so that M is semisimple. However, the thing I try to show M is simple but I am stuck there. Is there anyone can help me?

Let $M$ be co-isosimple and semi-hopfian, and $N$ be any submodule not equal to $M$.
Firstly of course, $M\cong M/N$. But that means the composition of the projection $\pi:M\to M/N$ with that isomorphism is an onto endomorphism of $M$. Therefore its kernel (which is $N$) is a summand of $M$. This shows that $M$ is a semisimple module.
If $M$ isn't finitely generated, then it is easy to come up with a submodule $N$ such that $M/N$ is finitely generated, and so $M\not\cong M/N$. Thus co-isosimplicity prevents $M$ from being infinitely generated.
So now suppose $M$ is finitely generated and has a nontrivial submodule $N$. Then the composition length of $M/N$ must be strictly smaller than the composition length of $M$, so $M\not\cong M/N$ in that case either. So co-isosimplicity precludes the existence of nontrivial submodules, too.
Therefore, the only thing left is for $M$ to be simple, which clearly works.
