Let $p,q$ be different primes and $D_{2pq^2}=\langle ab\mid a^{pq^2}=b^2=1, ba=a^{-1}b\rangle$ the dihedral group of order $2pq^2$. Find a composition series for $D_{2pq^2}$.
I don't know how to start this question. First, I have to find a maximal normal subgroup right? How can I do that? And how to proceed?
We can go step by step to construct a composition series.
It's easy to see that $\langle a \rangle$ is a maximal subgroup. It's order is $pq^2$, so it's index is $2$ and thus $\frac{D_{2pq^2}}{\langle a \rangle}$ is a simple group, which is enough to conclude that $\langle a \rangle $ is a maximal normal subgroup.
Now work in $\langle a \rangle \cong \mathbb{Z}_{pq^2}$. It is a cyclic group and thus the existence of subgroup of any order dividing $pq^2$ is guaranteed. For example you can take the subgroup $\langle a^p \rangle$ of order $q^2$ and index $p$. As it's quotent is $\mathbb{Z}_p$, simple group it's maximal normal subgroup of $\langle a \rangle$
Now simialrly $\langle a^{pq} \rangle$ is a maximal subgroup of $\langle a^{p} \rangle$. Finally $\langle a^{pq} \rangle \cong \mathbb{Z}_q$, which is a simple subgroup, so the composition series terminates with the trivial subgroup. Therefore we end up with the following composition series
$$\langle e \rangle \lhd \langle a^{pq} \rangle \lhd \langle a^{p} \rangle \lhd \langle a \rangle \lhd D_{2pq^2}$$