Upper central series members are strictly characteristic

Statement

Suppose $G$ is a group and $Z^{\alpha }(G)$ denote its $\alpha ^{th}$ 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 Strictly characteristic subgroup (?) of $G$.

Facts used

1. Center is strictly characteristic
2. Strict characteristicity is quotient-transitive
3. 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.