TITLE: Every normal subgroup of a finite group is contained in some composition series
QUESTION [12 upvotes]: In this context composition series means the same thing as defined here.
As the title says given a finite group $G$ and $H \unlhd G$ I would like to show there is a composition series containing $H.$
Following is my attempt at it.
The main argument of the claim is showing the following.

Lemma. If $H \unlhd G$ and $G/H$ is not simple then there exist a subgroup $I$ such that $H \unlhd I \unlhd G.$

The proof follows from the 4th isomorphism theorem since if $G/H$ is not simple then there is a normal subgroup $\overline{I} \unlhd G/H$ of the form $\overline{I} = H/I.$
Suppose now that $G/H$ is not simple. Using the above lemma we deduce that there exist a finite chain of groups (since $G$ is finite) such that $$H \unlhd I_1 \unlhd \cdots \unlhd I_k \unlhd G$$ and $G/I_k$ is simple. Now one has to repeat this process for all other pairs $I_{i+1}/I_{i}$ and for $I_1/H$ until the quotients are simple groups. This is all fine since all the subgroups are finite as well.
Now if $H$ is simple we are done otherwise there is a group $J \unlhd H$ and we inductively construct the composition for $H.$
Is the above "proof" correct? If so, is there a way to make it less messy?

REPLY [2 votes]: Suppose you have $A \triangleleft C$ with $C/A$ not simple. Then there is some nontrivial subgroup $B/A \triangleleft C/A$ (by Lattice Isomorphism Theorem), so you can extend $A \triangleleft C$ to $A \triangleleft B \triangleleft C$. By repeating this step you yield a composition series (it is clearly bounded, and will first terminate when all the composition factors are indeed simple).
Take $1 \triangleleft N \triangleleft G$ and apply above step. This constructs a composition series containing $N$.