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?