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I am trying to understand a proof to find the $A's$ fitting in the exact sequence $0 \rightarrow \Bbb Z_{p^m} \rightarrow A \rightarrow \Bbb Z_{p^n} \rightarrow 0$.

I have found an online solution, but there remains some problem to me: https://www3.nd.edu/~lnicolae/ProblemsHatcher.pdf

  1. How do we know that $p^ν g ∈ j(\Bbb Z_{p^m})$ here? I thought it may be done by some trials to find an element that is mapped to this element...But I have not find one.

Here is its position in the document

enter image description here

  1. Once we prove that $A$ is generated by the two elements $a_0$ and $a_1$, how does it necessarily imply the fact that $A\cong \Bbb Z_{p^\alpha}\oplus \Bbb Z_{p^\beta}$ with these conditions? I know that why $A$ contains a copy of $\Bbb Z_{p^m}$ so we need $\alpha\ge m$, but may I please ask why do we need $\alpha\ge n$ here?

Could someone please explain these two points? Thanks!

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  • $\begingroup$ I do not know the notation of Q1. But a natural choice would embed $Z_{p^m}$ to $Z_{p^m}\oplus Z_{p^n}$. Then project out $Z_{p^n}$. This is exact. The other way is to construct push-out by the exact sequence upstairs and a copy of $Z_{p^n}$ downstairs matching $Z_{p^n}$ top position with identity as morphism. $\endgroup$ – user45765 Sep 10 '17 at 13:03
  • $\begingroup$ @user45765 But may I please ask why we need $\alpha\ge n$ here? Also for the notation, I have edited to point out its position. I would appreciate if you would have a look at it. $\endgroup$ – PropositionX Sep 11 '17 at 1:54
  • $\begingroup$ I think the argument is for $\beta$. All you need is one of them greater than $m$ and the other one is greater than $n$. The argument is completely symmetric in terms of $\alpha$ and $\beta$ to me. I have not taken algebraic topology for long. So I do not think I will be able to supply a better explanation than algebraic version. I would suggest that you post this to algebraic topology or general topology part. $\endgroup$ – user45765 Sep 11 '17 at 2:22
  • $\begingroup$ @user45765 Thanks for your advice. I have edited it. It is indeed a problem in Hatcher's algebraic topology. But higely related to results from modules. $\endgroup$ – PropositionX Sep 12 '17 at 13:36
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In what follows I prefer to use the multiplicative notation for cyclic groups, the additive one being IMO too obfuscating for what really happens, so I'll note $C_r$ for the cyclic group of order $r$ with generator $a$ having elements $1,a,a^2,\ldots,a^{r-1}$. I'll start with an example where $p = 2$ and $p^m = 64, p^n = 8, p^\alpha = 128$ and $p^\beta = p^4$ :

EXAMPLE
Consider the exact sequence $C_{64} \rightarrow G \rightarrow C_8$. As shown in the referenced paper we have that one can take for example $G = C_{128} \times C_4$. Let $a, b, c$ be generators for $C_{64}, C_{128}, C_4$ respectively. Then the injection $j$ is given by $a \mapsto b^2c$. So the image of $C_{64}$ in G is the group $H$ with $64$ elements $1, b^2c, b^4c^2, b^6c^3\ldots b^{2i}c^i, \ldots$. We have to show that $G/H$ is cyclic of order $512/64 = 8$. Let us have a look at the cosets of $H$. The coset of $b$ is not trivial since it cannot be written as $b^{2i}c^i$. What are the values of $k$ such that $b^k \in H$. This is equivalent to asking: when is $b^k$ of the form $b^{2i}c^i$ ? Certainly the component in $c$ has to vanish, which happens only when $i = 0 \mod 4$ or equivalently $k = 0 \mod 8$. This proves that $G/H$ is cyclic of order $8$.

THE GENERAL CASE
Consider the exact sequence $C_{p^m} \rightarrow G \rightarrow C_{p^n}$. As shown in the referenced paper G has to be of the form $G = C_{p^{\alpha}} \times C_{p^\beta}$ with $\alpha+\beta = m + n$ and $\alpha >= \max(m,n,\beta)$. Let $a, b, c$ be generators for $C_{p^m}, C_{p^\alpha}, C_{p^\beta}$ respectively. Then the injection $j$ is given by $a \mapsto b^{p^{\alpha-m}}c$. So the image of $C_{p^m}$ in G is the group $H$ with $p^m$ elements $1, b^{p^{\alpha-m}}c, b^{2p^{\alpha-m}}c^2, b^{3p^{\alpha-m}}c^3\ldots b^{{ip^{\alpha-m}}}c^i, \ldots$. We have to show that $G/H$ is cyclic of order $p^{\alpha+\beta}/p^m = p^{\alpha+\beta-m} = p^n$. Let us have a look at the cosets of $H$. The coset of $b$ is not trivial since it cannot be written as $b^{{ip^{\alpha-m}}}c^i$. What are the values of $k$ such that $b^k \in H$. This is equivalent to asking: when is $b^k$ of the form $b^{{ip^{\alpha-m}}}c^i$ ? Certainly the component in $c$ has to vanish, which happens only when $i = 0 \mod p^\beta$ or equivalently $k = 0 \mod p^\beta p^{\alpha-m} $. This proves that $G/H$ is cyclic of order $p^{\alpha+\beta-m} = p^n$.

IMPORTANCE OF THIS QUESTION
The problem posed here is not only interesting in the domain of algebraic topology but also for people interested in finite groups. Indeed the notion of exact sequence is a generalization of the notion of (semi-direct) product and is known to pose certain problems (see extension of groups). Thats why I added another tag.

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  • $\begingroup$ Thanks for the detailed explaination! It is much neater and easier to understand then the referenced page. Now I think I get why all the $A's$ of the desired from fits into the exact sequence. But it seems to me that the question 2 remained unsolved. If possible, could you also give some for why we need all the $A's$ fits into the sequence is of this form? (the "only if" direction of the original problem). I am not quite understand the comment. $\endgroup$ – PropositionX Sep 13 '17 at 8:12
  • $\begingroup$ I know the structure theorem and think it is related, but it is unclear to me that once we show that $A$ is generated by the two elements, then we have $A\cong \Bbb Z_{p^\alpha}\oplus\Bbb Z_{p^\beta}$ with $\alpha+\beta=m+n,a\ge max\{m,n,\beta\}$. $\endgroup$ – PropositionX Sep 13 '17 at 8:19
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    $\begingroup$ It has nothing to do with groups but it is simply a trick with numbers. Think of any two numbers (I have $9$ and $13$ in mind because of the date today :-)) . If I chose two other numbers that have the same sum then one of them must be larger than each of the other ones (because elements of order $p^m$ and $p^n$ must exist in $G$). So in my example $(10, 12)$ and $(11,11)$ would not work, but $(9,13), (8,14),(7,15),\ldots,(0,22)$ do. The presence of $\beta$ in the $\max$ is simply because the larger of the two was labeled $\alpha$. $\endgroup$ – Marc Bogaerts Sep 13 '17 at 9:08

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