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I can't understand the proof which says that every finite abelian $p$-group can be written as a direct sum of cyclic $p$-groups.

I'm using Lang's book of Algebra. My problem is about the following preliminary LEMMA:

let $A$ be a finite abelian $p$-group. Let $a_1\in A$ be an element of maximal period, say $p^{r_1}$. Let $A_1$ be the cyclic subgroup of $A$ generated by $a_1$. We will use additive notation. So the LEMMA says: let $\overline{b}$ be an element of $A/A_1$ of period $p^r$, say. Then there exists a representative $a$ of $\overline{b}$ in $A$ which has the same period $p^r$.

$\textbf{proof}$ Let $b$ be any representative of $\overline{b}$ in $A$. Then $p^rb$ lies in $A_1$, say $p^rb=na_1$, for some $n\geq 0$. If $n=0$, we let $a=b$. So suppose $n\neq 0$. We note that the period of $\overline{b}$ is $\leq$ the period of $b$. Write $n=p^kt$ where $t$ is prime to $p$. Then $ta_1$ is also a generator of $A_1$, and hence has period $p^{r_1}$. We may assume $k\leq r_1$. Then $p^kta_1$ has period $p^{r_1-k}$. Then the lement $b$ has period $p^{r+r_1-k}$, whence by hypothesis, $r+r_1-k\leq r_1$ and $r\leq k$. $\textbf{This}$ 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 of $\overline{b}$ in $A$ and $p^ra=0$. Since period(a)$\geq p^r$ we conclude that period(a)=$p^r$.

$\textbf{Questions:}$

1) why $\textbf{this}$ proves the existence of $c$

2) where does $c$ come out from?

3) why period of a is greater equal than $p^r$?

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The element $c$ is constructed naturally: we have $$p^rb = na_1 = (p^kt)a_1=p^{k-r}p^rta_1.$$ If you put $c = p^{k-r}ta_1$, then it is clear that $c \in <a_1>$. Indeed we prove that $r \le k$ to guarantee that $p^{k-r}$ is an integer. For your third question, observe that $b + A_1 = a+ A_1$. If the period of a is $ u < p^r$ then $(ub + A_1) = (ua+ A_1)= 0$. That is, $\bar{b}^u = 0$ contradicting the fact that the period of $\bar{b}$ is $p^r$.

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