Prove $G$, $|G|=n$ is nilpotent $\iff$ $\forall m|n$, $G$ has a normal subgroup of order $m$. I got stuck in the second direction.
One direction: $|G|=n=p_1^{s_1}\cdot ...\cdot p_k^{s_k}$ Where $p_i$ prime. Particularly, $\forall p_i^{s_i}, p_i^{s_i}|n$ and therefore the Sylow-$p_i$ subgroup is unique and normal. Therefore every Sylow-$p$ subgroup is normal and that means $G$ is nilpotent. (There is a theorem\corollary claiming that saying every Sylow-$p$ subgroup is normal is equivalent to saying $G$ is nilpotent.
Other direction: Let $G$ be nilpotent. $|G|=n=p_1^{s_1}\cdot ...\cdot p_k1^{s_k}$ Where $p_i$ prime. Then, Sylow-$p_i$ subgroup is unique and normal. But what about the $p$-subgroups of order such as $p_i^{},p_i^{2},...,p_i^{s_i-1}$? They are subgroups contained in the Sylow-$p_i$ subgroup, and they all divide $n$, but are they normal? How can I show that?
Hints:
Since $\;G\;$ is nilpotent then every Sylow subgroup is normal. Now use the following:
Lemma 1: If $\;H\;$ is a finite $\;p$- group, $\;|H|=p^n\;$ , then for all $\;0\le k\le n\;$ there exists a normal subgroup $\;K\;$ of $\;H\;$ of order $\;p^k\;$
Proof (hints): use that $\;Z(H)\neq 1\;$, induction + the correspondence theorem (a subgroup of a quotient group is normal iff its inverse image under the quotient homomorphism is normal in the whole group)
Finally, as $\;G\;$ is the direct product of its Sylow subgroups and each pair of these commutes (why?) and etc.
Using the above (or whatever), you can also try the following nice
Lemma: The group $\;G\;$ is nilpotent iff $\;xy=yx\;$ for any pair of elements in the group with coprime orders.