Is the class of all nilpotent groups closed under homomorphic images? Suppose $H$ is a normal subgroup of a nilpotent group $G$. Does it imply, that  $\frac{G}{H}$ is nilpotent?
I do not know how to prove this. Could you help me, please?
 A: This is just the fourth isomorphism theorem, which says that if you have $N\trianglelefteq G$ and $N\le K\le G$ then the subgroups of $G/N$ are just the $K/N$. So take a central series, $\{A_i\}$, and project modulo $N$, and note that the condition $[G:A_i]\le A_i$ is satisfied on $[G/N: A_i/N]$ by construction. One might object and say that some of the $A_i$ of $G$ might be contained in $N$, but then that just produces a shorter central series for $G/N$, it doesn't change the facts for the ones that have $N\le A_i$.
A: If you are using upper central ($\zeta_i(G)$) and lower central series ($\gamma_{i+1}(G)$) a hint: $\gamma_{i+1}(G/H)=\gamma_{i+1}(G)H/H$ for every integer $i \geq 0$.
A: Subgroups and homomorphic images of nilpotent groups are nilpotent. The proof depends on your definition of a nilpotent group, which is usually by the upper or lower central series. In particular, quotients are nilpotent again, see here on MSE (including the comment of Qiaochu Yuan: "Nilpotence is defined in terms of certain words (iterated commutators) vanishing (having value the identity) identically on the group. Homomorphisms preserve this property").
A: Another approach and hint, if $G$ is finite and if $P \in Syl_p(G)$, then $P \unlhd G$. Since $H \unlhd G$, also $PH \unlhd G$ and hence $PH/H \unlhd G/H$, and $PH/H \in Syl_p(G/H)$.
