Let $A$ be a finite abelian $p$-group and $a_1\in A$ an element of maximal period. Let $A_1$ be the cyclic subgroup of $A$ generated by $a_1$, say of order $p^{r_1}$.Let $\bar{b}$ be an element of the factor group $A/A_1$, of period $p^r$. Then there exists a representative $a$ of $\bar{b}$ in $A$ which also has period $p^r$.
Proof. Let $b$ be any representative of $\bar{b}$ in $A$.Then $p^rb$ lies in $A_1$, say $p^r b=na_1$, with some integer $n\geq 0$. If $n=0$ let $a=b$. Suppose $n\neq 0$. We note that the period of $\bar{b}$ is $\leq$ the period of $b$. Write $n=p^k t$ with $t$ prime to $p$, then $ta_1$ is also a generator for $A_1$, hence has period $p^{r_1}$.We may assume $k\leq r_1$. Then $p^kta_1$ has period $p^{r_1-k}$. By our previous remark, the element $b$ has period $p^{r+r_1-k}$, whence by hypothesis, $r+r_1-k\leq r_1$ and $r\leq k$. This proves that there exists an element $c\in A_1$ such that $p^r c=p^r b$. Let $a=b-c$. Then $a$ is a representative of $\bar{b}$ in $A$ and $p^r a=0$. Since the period of $a$ is $\geq p^r$, we conclude that $a$ has period equal to $p^r$.
Can you help me understanding the bold sentence of the proof?