How to prove $A$ is generated by $a_0,a_1$?

Question

I am doing an exercise in Hatcher an stuck on it for a while.

This is: Finding the $A's$ fitting into the exact sequence:

$0 \rightarrow \Bbb Z_{p^m} \rightarrow A \rightarrow \Bbb Z_{p^n} \rightarrow 0$.

I am understanding the solution here:

https://www3.nd.edu/~lnicolae/ProblemsHatcher.pdf

And I have trouble understanding this part:

I think I need a complete formal proof for $A$ is generated by $a_0$ and $a_1$. To do this, I think I need to show that for all elements in $A$, it can be written as $ta_0+sa_1$ for some $t,s\in\Bbb Z$, conversely, any element of this form is in $A$. How can I prove it?

I do notice that there are three posts in this sites discussing this question. But all of them has not solved my problem yet. Could someone give some explaination at this point? Thanks.

Answer 1

Let $i:\Bbb Z_{p^m}\to A$ and $p:A\to \Bbb Z_{p^n}$ be the maps from the exact sequence. Let $a=i(1)\in A$, and $b$ be any element of $A$ with $p(b)=1$. If $c\in A$ then $p(c)=k\in\Bbb Z_{p^n}$ and let's think of $k$ as an integer. Then $p(c-kb)=0$, so $c-kb\in\ker p$. Thus $c-kp=i(u)$ with $u\in \Bbb Z_{p^m}$, thus $c-kb=la$ for some $l$.