I have a question about demonstrating part 2 of corollary 5.9.12 in Weibel's An Introduction to Homological Algebra.
Here is the setup. Fix a prime $p$ and suppose I have a long exact sequence of graded abelian groups \begin{equation} \cdots \to H_n \xrightarrow{\times p} H_n \to E_n \to H_{n-1} \to \cdots \end{equation} where each $H_n \to H_n$ map is given by multiplication by $p$. This rolls up to produce an exact couple \begin{equation} H_* \xrightarrow{i = \times p} H_* \xrightarrow{j} E_* \xrightarrow{k} H_* \end{equation} and thus an associated spectral sequence with $E^0_* = E_*$, called the Bockstein spectral sequence.
The statement in question is as follows:
(Weibel, cor. 5.9.12, page 159) In the Bockstein spectral sequence, an element $y \in H_n$ yields an element $j(y)$ of $E^r_n$ for all $r$; if $j(y) \neq 0$ in $E^{r-1}_n$ but $j(y) = 0$ in $E^r_n$, then $y$ generates a direct summand of $H_n$ isomorphic to $\mathbb{Z}/p^r$.
I'm having some trouble proving this. Ideally, I want some statement like "$j(y) = 0$ in $E^r_n$ if and only if $p^r y = 0$ in $H_n$". Anyway, if $j(y) = 0$ in $E^r_n$, then $j(y)$ is a $d^{r-1}$-boundary. Writing the differential explicitly, this amounts to \begin{equation} j(y) = d^{r-1} x = j(\frac{k^{r-1} x}{p^{r-1}}) \end{equation} for some $x \in E^{r-1}_{n+1}$.
Since $\ker j = p H_*$, this means that $y - \frac{k^{r-1} x}{p^{r-1}} = pz$ for some $z \in H_n$. Multiplying through by $p^r$ and using the fact that $p \cdot k^{r-1} x = 0$ by exactness, I find that $p^r y = p^{r+1} z$, or $p^r (y - p z) = 0$.
How do I conclude from this that $p^r y = 0$?
(Conversely, if $p^r y = 0$, then $p^{r-1} y \in \ker i^{r-1} = \operatorname{image} k^{r-1}$, and so $p^{r-1} y = k^{r-1} x$ for some $x$. In this case, $d^{r-1} x = j(y)$, so $j(y)$ is a $d^{r-1}$-boundary. This completes the proof of the claim.)
Also, why is $\langle y \rangle$ is a direct summand of $H_n$?
EDIT: I found a counterexample which I describe below in an answer. However, I would still like to know the "correct" statement of the corollary or a method to identify direct summands of $H_*$ using the Bockstein spectral sequence.
I suppose I must have misunderstood the statement, or am missing some crucial assumption, for I have found what I think is a counterexample to the corollary.
Consider the "long" exact sequence \begin{equation} 0 \rightarrow 0 \oplus \mathbb{Z}/p \rightarrow \mathbb{Z} \oplus \mathbb{Z}/p \xrightarrow{\times p} \mathbb{Z} \oplus \mathbb{Z}/p \rightarrow \mathbb{Z}/p \oplus \mathbb{Z}/p \rightarrow 0. \end{equation}
This rolls up to give the exact couple \begin{equation} \mathbb{Z} \oplus \mathbb{Z}/p \xrightarrow{i = \times p} \mathbb{Z} \oplus \mathbb{Z}/p \xrightarrow{j} \mathbb{Z}/p \oplus \mathbb{Z}/p \oplus \mathbb{Z}/p \xrightarrow{k} \mathbb{Z} \oplus \mathbb{Z}/p \end{equation}
If $y = (p, 1 + p^2 \mathbb{Z})$, then $j(y) = (0,0,1 + p \mathbb{Z})$, which is nonzero in $E^0$ but is zero is $E^1$. However, $y$ neither has order $p$ nor does it generate a direct summand of $\mathbb{Z} \oplus \mathbb{Z}/p$. Though of course, $y' = (0, 1 + p^2 \mathbb{Z})$ does satisfy the conclusions of the statement and $j(y') = j(y)$.