submodule of completely reducible module is completely reducible

Question

I attempted to solve problems from the textbook 'Basic Abstract Algebra' by P.B. Bhattacharya, S.K. Jain, and S.R. Nagpaul, specifically Chapter 14, Section 4, but encountered difficulties in completing some of them,

problem 14.4.1: Let $M$ be a completely reducible module, and let $K$ be a nonzero submodule of $M$. Show that $K$ is completely reducible. Also show that $K$ is a direct summand of $M$.

Here $$M=\oplus \Sigma_{\alpha \in \Lambda} M_\alpha$$ and $\exists \Lambda^{\prime} \subset \Lambda$ such that $M=K \oplus\left(\oplus \Sigma_{\alpha \in \Lambda'} M_\alpha\right)$.

$$ K\cong\frac{M}{\oplus \Sigma_{\alpha \in \Lambda^{\prime}} M_\alpha}=\frac{\oplus \Sigma_{\alpha \in \Lambda^{\prime \prime}} M_\alpha \oplus \Sigma_{\alpha \in \Lambda^{\prime}} M_\alpha}{\oplus \Sigma_{\alpha \in \Lambda^{\prime}} M_\alpha} \cong \sum_{\alpha \in \Lambda^{\prime \prime}} M_\alpha $$

Question 1: I didn't understand the above line specially the way $M$ is written using product of direct sum.

problem 14.4.3: Show that $\mathbb Z/(p_1 p_2)$ is a completely reducible $\mathbb Z$-module, where $p_1$ and $p_2$ are distinct primes.

Before I give a trial to this problem, I start to work with an example. Take $\mathbb Z_4$, then I assume this can be written as $\mathbb Z_2\oplus\mathbb Z_2$. but suspect $3\neq1+1$ can't be written as a direct sum here which seems not align with the definition. I feel that somewhere I need to use the fact that submodule of simple module $M$ is $(0)$ or $M$. I can't connect this with forward direction $(\implies)$.

Question 2: Any hint/solution for problem 14.4.3 will be appreciated.

Answer 1

Question 1: I didn't understand the above line specially the way is written using product of direct sum.

"Product of direct sum"? Where do you see that? I'm guessing you are referring to $\bigoplus$ versus $\sum$. There is no $\prod$ here.

I believe it is explicitly proven in this text that

The following are equivalent:

• $M$ is the sum of all of its simple submodules
• $M$ is a direct sum of some of its simple submodules.

So if you believe the hint you have written here that there exists such a set $\Lambda'$, then this demonstrates that $K$ is completely reducible.