If $NI am trying to understand this proof that if $N<M$ is a submodule, then $\ell(N)<\ell(M)$ (length of composition series):
Consider a submodule $N$ of $M$ where $N \neq M$. We first show
$$
\ell(N)<\ell(M) .
$$
Put $\ell=\ell(M)$ and let
$$
M=M_0 \supset M_1 \supset \ldots \supset M_{\ell}=\{0\}
$$
be a composition series of length $\ell$. Then
$$
N=N \cap M_0 \supseteq N \cap M_1 \supseteq \ldots \supseteq N \cap M_{\ell}=\{0\}
$$
is a chain of submodules from $N$ to $\{0\}$.
By a module isomorphism theorem, for each $i$,
$$
\begin{aligned}
\frac{N \cap M_i}{N \cap M_{i+1}} & =\frac{N \cap M_i}{\left(N \cap M_i\right) \cap M_{i+1}} \\
& \cong \frac{\left(N \cap M_i\right)+M_{i+1}}{M_{i+1}}
\end{aligned}
$$
the last of which is a submodule of the simple module $M_i / M_{i+1}$. Hence
$\frac{N \cap M_i}{N \cap M_{i+1}}$ is trivial or simple.
Thus, deleting repetitions from the above chain from $N$ to $\{0\}$ must yield a composition series for $N$, which proves
$$
\ell(N) \leq \ell(M) .
$$
I don't quite understand the statement "Thus, deleting repetitions from the above chain from $N$ to $\{0\}$ must yield a composition series for $N$". Why would that be the case?
 A: An example might clear things up I think: consider the two $\mathbb{Z}$-modules $M = \mathbb{Z}/(8)$ and its submodule $N \cong \mathbb{Z}/(4)$ generated by the element $2 \in \mathbb{Z}/(8)$. We have a composition series for $M$ given by $$
   M=M_0 = \mathbb{Z}/(8)\supseteq \underbrace{ M_1 = (2)}_{\cong \mathbb{Z}/(4)} \supseteq \underbrace{ M_2 = (4)}_{\cong \mathbb{Z}/(2)} \supseteq (0)
$$ which shows that $l(M)=2$ since each factor in the above series is isomorphic to $\mathbb{Z}/(2)$ which is indeed simple. Intersecting the above series with $N$ yields the series $$
N = (2) \cong \mathbb{Z}/(4)\supseteq \underbrace{ M_1 \cap N = N = (2)}_{\cong \mathbb{Z}/(4)} \supseteq \underbrace{ N\cap M_2 = M_2 = (4)}_{\cong \mathbb{Z}/(2)} \supseteq (0).
$$ As you mentioned in your argument, each factor in the above series must either be simple, as is the quotient $(M_1 \cap N)/(M_2 \cap N) \cong \mathbb{Z}/(2)$ for instance, or zero, like the first one is since $M \cap N = N = M_1 \cap N$. Thus, a composition series for $N$ can be obtained by simply removing one of each pair of successive terms with trivial quotient (i.e. successive terms which are in fact equal as submodules of $M$). $$
N \supseteq N \cap M_2 \supseteq (0) (\implies \ell(N) = 1)
$$
