Let $G$ be a reduced abelian primary group of lenght $\lambda$, and let $\alpha$ and $\beta$ with $\beta$ a limit ordinal and $\alpha < \beta \leq \lambda$. Show that the Ulm Invariants of $G_{\alpha}/G_{\beta}$ coincide with the Ulm Invariants of $G$ between $\alpha$ and $\beta$.
Question 35 from the book Infinite Abelian Groups by Irving Kaplansky, 1971 (second edition).
Definition$_0$: A group $G$ is divisible if for all $x\in G$ and all $n\in N$ there exists an $y \in G$ such that $x=ny$. If the only element that satisfies this property in $G$ is $0$ we call $G$ reduced.
A group $G$ is primary if for all $g\in G$ there exists an $n \in N$ such that $o(g)=p^n$ and $p$ is a prime number.
Definition$_1$: Let $G$ be a primary abelian group we define:
(i) $G_n=p^n G$
(ii) $G_{\omega}=\cap_{i\in N}G_n$
(iii) $G_{\omega+1}=pG_{\omega}$
(iv) If $\beta$ is a limit ordinal we define $G_{\beta}=\cap_{\alpha < \beta}G_{\alpha}$ (the intersection of all $\alpha<\beta$ where $\alpha$ is an ordinal), and $G_{\beta+1}=pG_{\beta}$.
(v) We define the group length $\lambda$ as the first ordinal such that $G_\lambda=G_{\lambda+1}$ (notice that this will be the maximal divisible part of the group, and if $G$ is reduced than $G_\lambda=0$).
Definition$_2$: Let $S$ be a subgroup of $G$, then we define $S_\alpha=G_\alpha \cap S$.
Definition (Ulm Invariants): Let $G$ be a group, we define $P=\{x\in G\mid px=0\}$. All the elements of $P$ have order $p$, so it's a vector space over $\mathbb F_p$. We define the $\alpha$-th Ulm Invariant of $G$ as the dimension of $P_\alpha/P_{\alpha+1}$ over $\mathbb F_p$.