Let $G$ be a group and $Z(G$) be the center of G. The upper central series
$1=Z_0(G)≤Z_1(G)≤…,$ is defined by $Z_{n+1}(G)/Z_n(G)=Z(G/Z_n(G))$.
Let $Z_n(G)$ be $n^{th}$ term of the upper central serie of a group $G$.
Prove that $Z_n(G)$ is a characteristic subgroup of $G.$
I use the math induction.
It is easy to prove that $Z_0(G)$ or $Z_1(G)$ a characteristic subgroup of $G.$ Assume that $Z_i(G) (i=0\ldots n-1)$ is a characteristic subgroup of $G.$ I can't deduce that $Z_i(G)$ is a characteristic subgroup of $G.$
Best Answer
Let $\alpha \in {\rm Aut}(G)$. By inductive hypothesis, $\alpha(Z_{n-1}(G)) = Z_{n-1}(G)$. So $\alpha$ induces an automorphism $\overline{\alpha}$ of $G/Z_{n-1}(G)$ defined by $\overline{\alpha}(gZ_{n-1}(G)) = \alpha(g)Z_{n-1}(G)$.
Since $Z(G/Z_{n-1}(G))$ is characteristic in $G/Z_{n-1}(G)$, $\overline{\alpha}$ fixes $Z(G/Z_{n-1}(G))$. But, by definition, $Z(G/Z_{n-1}(G)) = Z_n(G)/Z_{n-1}(G)$ so, for any $g \in Z_n(G)$, $\overline{\alpha}(gZ_{n-1}(G)) = \alpha(g)Z_{n-1}(G)\in Z_n(G)/Z_{n-1}(G)$, and hence $\alpha(g) \in Z_n(G)$.
We have proved that $\alpha(Z_n(G)) \le Z_n(G)$ Similarly $\alpha^{-1}(Z_n(G)) \le Z_n(G)$ and applying $\alpha$ to this gives $Z_n(G)) \le \alpha(Z_n(G))$. Hence $Z_n(G)$ is characteristic in $G$.