Need help proving any subgroup and quotient of a nilpotent group is nilpotent? $G$ is the direct product of its Sylow subgroups $P_i$. Then if $H \le G$, $P_i \cap H \le H$. I'm stuck now on how to proceede to find a terminating central series for $H$.
Also, I know $P_i H / H \le G/H$ so I'm assuming proving the above will prove that the quotient is nilpotent too?

I posted a proof that a quotient of a nilpotent group is nilpotent recently.
Here is a proof that a subgroup of a nilpotent group is nilpotent.
I will work with the definition that a group $G$ is nilpotent if it has a central series
$$1 = N_0 \leq N_1 \leq \cdots \leq N_r = G$$
where each $N_i \lhd G$ and $N_i / N_{i-1} \leq Z(G/N_{i-1})$ for each $i$.
Suppose that $G$ is nilpotent and $H \leq G$. We wish to show that $H$ is nilpotent. Define $M_i = H \cap N_i$ for $0 \leq i \leq r$. Note that $M_0 = H \cap N_0 = H \cap 1 = 1$, and $M_r = H \cap N_r = H \cap G = H$. Also, $H \cap N_{i} \lhd H$ by the diamond isomorphism theorem.
It remains to show that
$$M_i / M_{i-1} \leq Z(H / M_{i-1})$$
Once again by the diamond isomorphism theorem, we have an isomorphism $\phi :  H / M_{i-1} \to HN_{i-1}/N_{i-1}$ given by $\phi(hM_{i-1}) = hN_{i-1}$. The image of $M_i / M_{i-1}$ is $\phi(M_i / M_{i-1}) = M_{i}N_{i-1}/N_{i-1}$.
Since $M_{i} = H \cap N_{i} \leq N_{i}$ and $N_{i-1} \leq N_{i}$, it follows that $M_{i}N_{i-1}/N_{i-1} \leq N_i / N_{i-1}$, so $\phi(M_{i} / M_{i-1}) \leq N_i / N_{i-1} \leq Z(G/N_{i-1})$.
Therefore,
$$\begin{aligned}
\phi(M_i/M_{i-1}) &\leq (HN_{i-1} / N_{i-1}) \cap Z(G/N_{i-1}) \\
&\leq Z(HN_{i-1} / N_{i-1})\\
& = Z(\phi(H/M_{i-1}))\\
&= \phi(Z(H/M_{i-1}))\\
\end{aligned}$$
where the last equality follows because $\phi$ is an isomorphism, so the center is preserved by $\phi$.
Finally, again since $\phi$ is an isomorphism, we can apply $\phi^{-1}$ to both sides to conclude that $M_i / M_{i-1} \leq Z(H/M_{i-1})$, as desired.
