A finite $\mathbb{Z}$-module whose submodules are totally ordered by inclusion.

Question

I have the following problem:

Let $M$ be a finite $\mathbb{Z}$-module such that set of the submodules is totally ordered by inclusion. Prove that there exist a prime $p$ such that $|M|=p^\alpha$ for some $\alpha$.

I'd like some hint to solve this problem.

Thanks a lot!!!

Answer 1

Ok so a finite $\mathbb{Z}$ module is a finite abelian group ! If its order is divisible by two different primes then there are 2 elements of relatively prime orders, and their subgroups cannot have an inclusion one in the other.

Answer 2

Hint: let $L_1$ be the unique minimal submodule; then $L$ must be simple, so…

Let $L_2/L_1$ be the unique minimal submodule of $M/L_1$, so…