suppose $G$ is a finite group, $G$ is nilpotent iff every quotient group of $G$ has no trivial center.

Question

suppose $G$ is a finite group, $G$ is nilpotent iff every quotient group of $G$ has no trivial center.

any hint or guidance will be great,I badly stuck in this one,thanks.

Answer 1

Following Jack's hint: the upper central series

$$1=: Z_0\le Z_1\le\ldots\le Z_m=G\;,\;\;\text{with}\;\;Z_i/Z_{i-1}=Z\left(G/Z_{i-1}\right)\;,\;\;z=1,...,m$$

We need thus to prove that there exists $\;x\in G\;\;s.t.\;\;[x,g]\in N\;\;\forall\,g\in G\;$

Assume $\;N\neq 1\,,\,G\;$ , and take the maximal index

$$\;0< i_0< m\;\;\;\text{s.t.}\;\; \;Z_{i_0}\le N\;\;\;\text{(and thus}\;Z_{i_0+1}\lneqq N\,)$$

(observe that $\;N\neq 1\,,\,G\implies 0<i_0<m\;$ , since $\;N\cap Z(G)\neq 1\;$) , and then

$$\forall x\in G\;,\;\;xZ_{i_0}\in \overbrace{Z\left(G/Z_{i_0}\right)}^{=Z_{i_0+1}/Z_{i_0}}\implies [x,g]\in Z_{i_0}\le N\;\;\forall \,g\in G\implies Z(G/N)\neq 1\;$$

and note that we can now choose $\;x\notin N\;$ ...

Answer 2

Another proof goes by induction on the order. By the hypothesis $Z(G) \ne \{1\}$, and then by the inductive assumption $G/Z(G)$ is nilpotent, as a quotient of $G/Z(G)$ is also a quotient of $G$.

Thus for some $n$ we have $$ \gamma_{n}\left(\frac{G}{Z(G)}\right) = \frac{\gamma_{n}(G) Z(G)}{Z(G)} = \frac{Z(G)}{Z(G)}, $$ so $\gamma_{n}(G) \le Z(G)$ and $\gamma_{n+1}(G) = \{ 1 \}$.

Here $\gamma_{n}(G)$ is the $n$-th term of the lower central series.