Finding all abelian groups such that there exists certain short exact sequence.

Question

I have to find all abelian groups $A$ such that there exists a short exact sequence $0\rightarrow\mathbb{Z}\rightarrow A\rightarrow\mathbb{Z}\oplus\mathbb{Z}_{6}\rightarrow 0$.

I have found $A=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_{6}$, $A=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_{2}$, $A=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_{3}$, and $A=\mathbb{Z}\oplus\mathbb{Z}$. I am not able to find any other, but I am also not able to prove these are the only ones.

This exercise has come up in an Algebraic Topology course, so I would like to prove it using elementary abstract algebra and not Homological Algebra.

Any hint will be greatly appreciated.

Answer 1

If you want to avoid using ${\rm Ext}$, I think here is a possible plan.

Given an exact sequence $0\rightarrow \mathbb{Z}\rightarrow A\rightarrow \mathbb{Z}\oplus \mathbb{Z}/6\mathbb{Z}$, we can form the following commutative diagram: $$\require{AMScd} \begin{CD} 0 @>>> \mathbb{Z} @>>> \mathbb{Z}^3 @>>> \mathbb{Z}^2 @>>> 0 \\ @. @| @VVV @VV V @. \\ 0 @>>> \mathbb{Z} @>>> A @>>> \mathbb{Z}\oplus \mathbb{Z}/6\mathbb{Z} @>>> 0 \end{CD} $$ The idea is to present $A$ as a surjection $\mathbb{Z}^3\rightarrow A$ and compute the kernel.

You can do this by taking elements and computing maps by hand, but for clarity, we can extend the diagram to $$ \begin{CD} @. 0 @>>>0 @>>>0 @. \\ @. @VVV @VVV @VV V @. \\ 0 @>>> 0 @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> 0 \\ @. @VVV @VVV @VV V @. \\ 0 @>>> \mathbb{Z} @>>> \mathbb{Z}^3 @>>> \mathbb{Z}^2 @>>> 0 \\ @. @| @VVV @VV V @. \\ 0 @>>> \mathbb{Z} @>>> A @>>> \mathbb{Z}\oplus \mathbb{Z}/6\mathbb{Z} @>>> 0\\ @. @VVV @VVV @VV V @. \\ @. 0 @>>>0 @>>>0 @. \\ \end{CD} $$ We now focus on the upper right square:

$\require{AMScd} \begin{CD} \mathbb{Z} @>>> \mathbb{Z} \\ @VVV @VVV\\ \mathbb{Z}^3 @>>> \mathbb{Z}^2 \\ \end{CD} $

The right vertical map is $1\rightarrow \begin{pmatrix} 6 \\ 0\end{pmatrix}$. Since the diagram commutes, the left vertical map is $1\rightarrow \begin{pmatrix} a \\ 6 \\ 0\end{pmatrix}$, where $a$ is some undetermined integer.

Finally, this means $A$ is the cokernel of the exact sequence \begin{align*} 0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}^3\rightarrow A\rightarrow 0, \end{align*} where $\mathbb{Z}\rightarrow \mathbb{Z}^3$ is given by $\begin{pmatrix} a \\ 6 \\ 0\end{pmatrix}$. By taking row operations, we see that $A\cong \mathbb{Z}^2\oplus \mathbb{Z}/{\rm gcd}(a,6)\mathbb{Z}$, so I think your list is complete.

Edit: To be clear, there are only 4 groups for $A$ on this list, while the Ext group is apparently order 6. This is because Ext also keeps track of the maps in the short exact sequence $0\rightarrow \mathbb{Z}\rightarrow A\rightarrow \mathbb{Z}\oplus \mathbb{Z}/6\mathbb{Z}\rightarrow 0$, and not just the isomorphism class of $A$.

Remark 2: Computing the kernel of $\mathbb{Z}^3\rightarrow A$ is the most technical part, but I didn't realize, as applying the 9-lemma was automatic. The proof of the 9-lemma is a diagram chase, though it is slightly easier in our case. Unfortunately, a lot of the objects in the big diagram are $\mathbb{Z}$'s, which is confusing when I can't point to things, so consider:

$$ \begin{CD} @. 0 @. 0 @. 0 @. \\ @. @VVV @VVV @VV V @. \\ 0 @>>> A_{11} @>>> A_{12} @>>> A_{13} @>>> 0 \\ @. @VVV @VVV @VV V @. \\ 0 @>>> A_{21} @>>> A_{22} @>>> A_{23} @>>> 0 \\ @. @VVV @VVV @VV V @. \\ 0 @>>> A_{31} @>>> A_{32} @>>> A_{33} @>>> 0\\ @. @VVV @VVV @VV V @. \\ @. 0 @. 0 @. 0 @. \\ \end{CD} $$

Here, $A_{21}=\mathbb{Z}$, $A_{22}=\mathbb{Z}^3$, $A_{32}=\mathbb{Z}^2$, etc. What we need to determine is $A_{11}$, $A_{12}$, and $A_{13}$ (the top row).

1) $A_{11}$ is the kernel of $\mathbb{Z}\rightarrow\mathbb{Z}$, so it is 0.

2) $A_{13}$ is the kernel of $\mathbb{Z}^2\rightarrow \mathbb{Z}\oplus \mathbb{Z}/6$, so it is $\mathbb{Z}$.

3) We want $A_{12}$. There are induced maps from $A_{11}$ to $A_{12}$ and $A_{12}$ to $A_{13}$ that make the diagram commute. For example, if $a\in A_{12}$, then $a$ maps to 0 under $A_{12}\rightarrow A_{22}\rightarrow A_{23}\rightarrow A_{33}$, which means $a$ maps to ${\rm ker}(A_{23}\rightarrow A_{33})$ under $A_{12}\rightarrow A_{22}\rightarrow A_{23}$. Since $A_{13}={\rm ker}(A_{23}\rightarrow A_{33})$ this means $a$ actually maps to $A_{13}$.

4) From the comments, we also know that the map $A_{12}\rightarrow A_{13}$ is surjective. We want to show that it has no kernel. Suppose $a\in {\rm ker}(A_{12}\rightarrow A_{13})$. Then, $a$ maps to 0 under $A_{12}\rightarrow A_{22}\rightarrow A_{23}$, which means $a$ maps to ${\rm ker}(A_{22}\rightarrow A_{23})\subset A_{22}$ under $A_{12}\rightarrow A_{22}$. So, we can regard the image of $a$ in $A_{22}$ as $b$, where $b$ is an element of $A_{21}$.

5) We know that $b$ maps to 0 in $A_{31}$ as $A_{31}\rightarrow A_{32}$ is injective, and $A_{21}\rightarrow A_{31}\rightarrow A_{32}$ is the same as $A_{21}\rightarrow A_{22}\rightarrow A_{32}$, and the image of $b$ under $A_{21}\rightarrow A_{22}\rightarrow A_{32}$ is the same as the image of $a$ under $A_{12}\rightarrow A_{22}\rightarrow A_{23}$, which is 0.

6) But the kernel of $A_{31}\rightarrow A_{32}$ is 0. This means $b=0$. Since $A_{12}\rightarrow A_{22}$ is injective, $a=0$. This means $A_{12}\rightarrow A_{13}$ is a bijection, so $A_{13}=\mathbb{Z}$.

You might want to consider looking up how to prove snake lemma (http://mathworld.wolfram.com/SnakeLemma.html), and then directly apply that to prove the 9-lemma, since you mentioned you're taking an algebraic topology course, which is going to use this in the future anyways.

Answer 2

Partial Answer: Let $R$ be a commutative ring with identity and let $0 \rightarrow M \rightarrow E \rightarrow N \rightarrow 0$ be an exact sequence of $R$ modules. Then the equivalence class of extensions is given by ${\rm Ext}^1_{R}(N, M).$ (See any Homological Algebra book.)

In this case $R = \mathbb{Z}, M = \mathbb{Z}, N = \mathbb{Z} \oplus \mathbb{Z}_6.$ So need to look at the abelian group ${\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z} \oplus \mathbb{Z}_6, \mathbb{Z}).$ First of all, ${\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z} \oplus \mathbb{Z}_6, \mathbb{Z}) \cong {\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Z}) \oplus{\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z}_6, \mathbb{Z}).$ Now ${\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Z}) = 0$ and ${\rm Ext}^1_{\mathbb{Z}}(\mathbb{Z}_6, \mathbb{Z}) \cong \mathbb{Z}_6$. (See any Homological Algebra book for this results and how to prove it.)

So we have exactly $6$ non-isomorphic choice for $A$. (But I don't know how to construct them all.)

Answer 3

The extensions by $\def\Z{\mathbb{Z}}\Z\oplus\Z_6$ of $\Z$ are described by the group $\def\Ext{\operatorname{Ext}}\Ext(\Z\oplus\Z_6,\Z)$. By general theory, this is isomorphic to $\Ext(\Z,\Z)\oplus\Ext(\Z_6,\Z)$. As the first summand is zero, because $\Z$ is free, we need to compute the second group; this can be done by considering the exact sequence $$ 0\to\Z\xrightarrow{\cdot6}\Z\to\Z_6\to0 $$ (where $\cdot6$ denotes “multiplication by $6$) and applying to it the contravariant functor $\def\Hom{\operatorname{Hom}}\Hom(\Z,-)$: $$ 0\to\Hom(\Z_6,\Z)\to\Hom(\Z,\Z)\xrightarrow{\cdot6}\Hom(\Z,\Z)\to \Ext(\Z_6,\Z)\to\Ext(\Z,\Z) $$ which can be rewritten as $$ 0\to0\to\Z\xrightarrow{\cdot6}\Z\to\Ext(\Z_6,\Z)\to0 $$ which means that $\Ext(\Z_6,\Z)\cong\Z_6$ and that the isomorphism is exactly the same as the “natural” isomorphism $\Z_6\cong\Hom(\Z_6,\Z_6)$.

In particular there are six non-equivalent extensions of the kind you're looking for. How can you write them?

First, if you're given $0\to\Z\to A\to\Z\oplus\Z_6\to0$, you can say that $\Z$ is a homomorphic image of $A$ and split it off; so $A\cong B\oplus\Z$ and we have an exact sequence $0\to\Z\to B\to\Z_6\to0$. This corresponds to $$ \Ext(\Z\oplus\Z_6,\Z)\cong\Ext(\Z,\Z)\oplus\Ext(\Z_6,\Z)=0\oplus\Ext(\Z_6,\Z). $$ Now, let $x\in\Z_6\cong\Hom(\Z_6,\Z_6)$; this corresponds to the endomorphism $\cdot x\colon\Z_6\to\Z_6$ (the multiplication by $x$), so you can consider a pull-back diagram $$\require{AMScd} \begin{CD} 0 @>>> \Z @>>> B @>>> \Z_6 @>>> 0 \\ @. @| @VVV @VV{\cdot x}V @. \\ 0 @>>> \Z @>{\cdot6}>> \Z @>>> \Z_6 @>>> 0 \end{CD} $$ It can be proved that, as $x\in\Z_6$, these are the inequivalent extensions, but this would require quite a long argument, that you find in any book dealing with homological algebra.

The “split extension”, $B=\Z\oplus\Z_6$, that is the same as the one you found, corresponds to taking $x=0$.

The pullback $B$ is the subgroup of $\Z\oplus\Z_6$ consisting of the pairs $(a,b)$ such that $a+6\Z=xb$. In the case of $x=0$ it's just $6\Z\oplus\Z_6\cong\Z\oplus\Z_6$. In the case of $x=1$ it is the set of pairs $(a,b)$ such that $a+6\Z=b$. Do the same for the remaining four endomorphisms of $\Z_6$.

Answer 4

Hint: For group extensions in a short exact sequence $$0\to A\to E\to B\to 0$$ for each of the homomorphisms $B\to{\rm Out}A$ there is an extension $E$.

Also ${\rm Out}\ \Bbb{Z}={\Bbb{Z}}_2$.