Upper central series members are strictly characteristic

Statement
Suppose $$G$$ is a group and $$Z^\alpha(G)$$ denote its $$\alpha^{th}$$ fact about::upper central series member. In other words:


 * $$Z^0(G)$$ is the trivial subgroup.
 * When $$\alpha = \beta + 1$$ is a successor ordinal, $$Z^{\alpha}(G)$$ is the inverse image of the center of $$G/Z^{\beta}(G)$$ with respect to the natural projection map $$G \to G/Z^{\beta}(G)$$. In other words, $$\! Z^\alpha(G)/Z^\beta(G) = Z(G/Z^\beta(G))$$.
 * When $$\alpha$$ is a limit ordinal, $$Z^{\alpha}(G)$$ is the union of all $$Z^{\gamma}(G)$$ where $$\gamma < \alpha$$.

Then, each $$Z^\alpha(G)$$ is a fact about::strictly characteristic subgroup of $$G$$.

Facts used

 * 1) uses::Center is strictly characteristic
 * 2) uses::Strict characteristicity is quotient-transitive
 * 3) uses::Strict characteristicity is strongly join-closed

Proof
We prove the result by transfinite induction on $$\alpha$$.


 * 1) Base case $$\alpha = 0$$: The trivial subgroup is strictly characteristic.
 * 2) Case where $$\alpha = \beta + 1$$, a successor ordinal: By the inductive hypothesis, $$Z^\beta(G)$$ is strictly characteristic. Further, $$Z^\alpha(G)/Z^\beta(G) = Z(G/Z^\beta(G))$$. By fact (1), this is a strictly characteristic subgroup of $$G/Z^\beta(G)$$. By fact (2), we obtain that $$Z^\alpha(G)$$ is strictly characteristic in $$G$$.
 * 3) Case where $$\alpha$$ is a limit ordinal: $$Z^\alpha(G)$$ is the union of $$Z^\gamma(G)$$, $$\gamma < \alpha$$. By the inductive assumption, the $$Z^\gamma(G)$$ are all strictly characteristic. So, by fact (3), $$Z^\alpha(G)$$ is strictly characteristic.