What is a composition series for $N$ and $M/N$ 
Let $M$ be a module with a composition series. Show that every submodule of $M$ is a member of a composition series for $M$.

I understand the concept of a composition series for a module: it's a chain of submodules that has the empty set on one end and the module itself on the other end, and every submodule is a maximal submodule of the next one in the chain.
The answer my book (Introduction to Ring Theory, Paul Cohn) gives is:

Given $N\subset M$, take composition series of $N$ and $M/N$.

I understand what a quotient group is, so I'm not asking what $M/N$ means. My question is: what does it mean to take a composition series of two things?
 A: Let $0 = X_0 \subseteq X_1 \subseteq X_2 \subseteq \cdots \subseteq X_a = N$ be a composition series for $N$. Let $0 = Y_0 \subseteq Y_1 \subseteq Y_2 \subseteq \cdots \subseteq Y_b = M/N$ be a composition series for $M/N$. This gives rise to a sequence $N = Z_0 \subseteq Z_1 \subseteq Z_2 \subseteq \cdots \subseteq Z_b = M$ of submodules of $M$ defined by $Z_i = \pi^{-1}\left(Y_i\right)$, where $\pi$ is the canonical projection $M \to M/N$. It satisfies
(1) $Z_i / Z_{i-1} \cong Y_i / Y_{i-1}$ for every $i \in \left\{1,2,\ldots,b\right\}$
(check this!).
Now, "glueing together" the two sequences $0 = X_0 \subseteq X_1 \subseteq X_2 \subseteq \cdots \subseteq X_a = N$ and $N = Z_0 \subseteq Z_1 \subseteq Z_2 \subseteq \cdots \subseteq Z_b = M$, we obtain a new sequence
$0 = X_0 \subseteq X_1 \subseteq X_2 \subseteq \cdots \subseteq X_a = N = Z_0 \subseteq Z_1 \subseteq Z_2 \subseteq \cdots \subseteq Z_b = M$.
We can write it in a more homogeneous way: Namely,
(2) $0 = P_0 \subseteq P_1 \subseteq P_2 \subseteq \cdots \subseteq P_{a+b} = M$,
where for each $i \in \left\{0,1,\ldots,a+b\right\}$, we define the submodule $P_i$ of $M$ as follows: If $i \leq a$, then set $P_i = X_i$; else, set $P_i = Z_{i-a}$.
Notice that
(3) $P_i = Z_{i-a}$ for every $i \in \left\{a,a+1,\ldots,a+b\right\}$.
Indeed, (3) follows directly from the definition of $P_i$ when $i \neq a$; otherwise, it follows from the computation
$P_i = P_a$ (since $i=a$)
$= X_a = N = Z_0 = Z_{i-a}$ (since $i=a$ and thus $0=i-a$).
Now, I claim that (2) is a composition series for $M$. In order to prove this, we need to show that for every $i \in \left\{1,2,\ldots,a+b\right\}$, the quotient $P_i/P_{i-1}$ is a simple module.
So let $i \in \left\{1,2,\ldots,a+b\right\}$. We need to show that $P_i/P_{i-1}$ is a simple module.
If $i \leq a$, then $P_i/P_{i-1} = X_i/X_{i-1}$ (since $P_i=X_i$ and $P_{i-1}=X_{i-1}$) is a simple module (since $0 = X_0 \subseteq X_1 \subseteq X_2 \subseteq \cdots \subseteq X_a = N$ is a composition series), and so we are done in this case. Hence, for the rest of this proof, we WLOG assume that $i > a$. Then, (3) yields $P_i = Z_{i-a}$. Also, (3) (applied to $i-1$ instead of $i$) yields $P_{i-1} = Z_{i-1-a} = Z_{i-a-1}$. Hence, $P_i / P_{i-1} = Z_{i-a} / Z_{i-a-1} \cong Y_{i-a} / Y_{i-a-1}$ (by (1), applied to $i-a$ instead of $i$) is a simple module (since $0 = Y_0 \subseteq Y_1 \subseteq Y_2 \subseteq \cdots \subseteq Y_b = M/N$ is a composition series). Thus, our proof is complete.
