The following is a lemma in Algebra (page 44): $A$ is a finite abelian p-group. Let $\overline{b}$ be an element of $A/A_1$ ($A_1$ is a cyclic group generated by $a_1 \in A$ of period $p^{r_1}$), of period $p^r$. Then there exists a representative a of $\overline{b}$ in $A$ which also has period $p^r$.
Proof. Let b be any representative of $\overline{b}$ in A. Then $p^rb$ lies in A, say $p^rb = na$ with some integer $n \ge 0$. We note that the period of $\overline{b}$ is $\le$ the period of b. If $n = 0$ we are done.Otherwise write $n = p^k\mu$, where $\mu$ is prime to $p$. Then $\mu a_1$ is also a generator of $A_1$, and hence has period $p^{r_1}$. We may assume $k \le r_1$. Then $p^k\mu a_1$has period $p^{r_1-k}$. By our previous remarks, the element $b$ has period $$p^{r+r_1-k}$$ whence by hypothesis,$\underline{r + r_1 - k \le r_1}$ and $r \le k$. This proves that there exists an element $c\in A_1$ such that $p^rb = p^rc$. Let $a = b - c$. Then $a$ is a representative for $\overline{b}$ in $A$ and $p^ra = 0$. Since period $(a) < p^r$ we conclude that $a$ has period equal to $p^r$ .
My question is, how the underline part in the proof makes sense.
Let's get this off the unanswered queue...
There is a hypothesis that is not stated by the OP, which is that $a_1$ has maximal period in $A$. That is, for all $b\in A$, $|b|\leq |a_1|=p^{r_1}$.
Since we have that $b^{p^r}\in \langle a_1\rangle$, it has order $p^{r_1-k}$ for some $k$, $0\leq k\leq r_1$. This means that $b$ has order precisely $p^{r+(r_1-k)}$. Since $|b|\leq p^{r_1}$, it follows that $p^{r+r_1-k}\leq p^{r_1}$, which yields the underlined inequality, $r+r_1-k\leq r_1$.