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
- 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
- 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!
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.