Hatcher exercise 2.1.14

Question

Determine which abelian groups $A$ fit into a short exact sequence $0\to\Bbb Z_{p^m}\xrightarrow{f} A\xrightarrow{g}\Bbb Z_{p^n}\to 0$ with $p$ prime.

I already showed that if $A=\Bbb Z_{p^{m+n-k}}\ \oplus\Bbb Z_{p^k}$ for $0\leq k\leq\min\{m,n\}$ the sequence is exact. Left to show any such $A$ is of that form. For that, I showed that $A$ is generated by two elements $\alpha,\beta$ such that $f(1) = \alpha$ and $g(\beta)=1$. And also by the fundamental theorem of finitely generated abelian group, $A\simeq \Bbb Z_{p^{m_1}}\oplus\cdots\oplus\Bbb Z_{p^{m_h}}$ for some $h\geq 1$ and $m_i\geq 0$. But I can't conclude $A\simeq Z_{p^{m+n-k}}\ \oplus\Bbb Z_{p^k}$. Could you help?

What you write is incorrect. If $k\neq n$, then $\alpha$ has order strictly larger than $p^m$, and so the map $f$ you describe is not well defined, since $f(p^m)$ would equal $p^m\neq 0$, but $p^m=0$ in $\mathbb{Z}_{p^m}$. — Arturo Magidin, May 10 '21 at 18:33
@ArturoMagidin I defined a map $f$ by $1\mapsto (p^{n-k},1)$. Then the order of $\operatorname{Im}(f)$ is $p^m$ — one potato two potato, May 10 '21 at 18:40
That's not what you wrote in your post. Are we supposed to read your mind? — Arturo Magidin, May 10 '21 at 18:41
@ArturoMagidin Those $f,g$ in the post is an arbitrary map that makes the sequence exact. What I gave is the case when $A =\Bbb Z_{p^{m+n-k}}\ \oplus\Bbb Z_{p^k}$ which I omitted and state 'I showed' — one potato two potato, May 10 '21 at 18:44
Sorry; what you wrote is confusing to me. If you know that $A$ can be generated by two elements, then you know that $h\leq 2$. And you know that $m_1\geq m$ (assuming you are summands in nonincreasing size). So what are you missing? — Arturo Magidin, May 10 '21 at 18:49
@ArturoMagidin Then as $m_1+m_2 = m+n$, $0\leq m_2\leq \min{m,n}$ right? — one potato two potato, May 10 '21 at 19:00
$m_1+m_2=m+n$, $m_1\geq m$, so $m_1-m = n-m_2\geq 0$; that gives $m_2\leq n$, and $m_2\leq m_1\leq m$ gives $m_2\leq m$. — Arturo Magidin, May 10 '21 at 19:01
@ArturoMagidin Wait, there is an error in your comment. $m_2\leq m_1\geq m$ not $m_2\leq m_1\leq m$. — one potato two potato, May 10 '21 at 19:39
Oops; quite right. Since $m_1\geq \max(m,n)$, $m+n=m_1+m_2$; so $m_1-\max(m,n) = \min(m,n)-m_2\geq 0$. — Arturo Magidin, May 10 '21 at 19:51
@ArturoMagidin Why there exists an element of order at least $p^{\max(m,n)}$? — one potato two potato, May 10 '21 at 19:59
Because you must have an element of order at least $p^m$ to be able to embed $\mathbb{Z}{p^m}$, and you must have an element of order at least $p^n$ to be able to project onto $\mathbb{Z}{p^n}$. — Arturo Magidin, May 10 '21 at 19:59

score 0 · Answer 1 · answered May 10 '21 at 18:36

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If $A$ is generated by two elements, it must be isomorphic to the direct sum of two cyclic groups, whose order sums to $p^{m+n}$. Up to isomorphism, the class of groups you described contains all these groups.

answered May 10 '21 at 18:36

R. Maresca

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"Generated by two elements" does not mean "minimally generated by two elements". The cyclic group is generated by two elements. And not every group described fits into an exact sequence as given. – Arturo Magidin May 10 '21 at 18:36

Hatcher exercise 2.1.14

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