Subgroup of polycyclic group is a polycyclic subgroup By definition $G$ is polycyclic if there exists a normal series where all its factors are cyclic, that's to say:
$\exists\ \{e\}=G_0 \unlhd G_1\unlhd\ldots\unlhd G_{n-1}\unlhd G_n=G $ with $\frac{G_k}{G_{k-1}}$ cyclic for all $k$ with $1\leq k\leq n$.
Let $H\leqslant G$ then proof $H$ is polycyclic.
What I have tried is taking $H_i =G_i \cap H$ for all $i$ with $0 \leq i\leq n $. We have that all $H_i$ are subgroup of $G$ and $G_{k-1}\unlhd G_k$ implies $H_{k-1}\unlhd H_k$.
Now we have to proof that $\frac{H_k}{H_{k-1}}$ is cyclic. I'm trying to show that  $\frac{H_k}{H_{k-1}} \leqslant \frac{G_k}{G_{k-1}}$ because $\frac{G_k}{G_{k-1}}$ is cyclic and that would be the proof. But I don't know how to proof that.
Searching in the net I've found that $\frac{H_k}{H_{k-1}} \cong \frac{G_{k-1}H_k}{G_{k-1}}\leqslant \frac{G_k}{G_{k-1}}$, but I'm not understanding it. Any help would be appreciated to understand that final sentence.

$$\begin{align}
\frac{H_k}{H_{k-1}}&=\frac{G_k\cap H}{G_{k-1}\cap H}\\
&=\dfrac{G_k\cap H}{G_{k-1}\cap(G_{k}\cap H)}\\
&\cong \dfrac{(G_k\cap H)G_{k-1}}{G_{k-1}}\\
&\leq \dfrac{G_k}{G_{k-1}} .
\end{align}$$
