TITLE: Stanford math qual: Abelian groups $G$ satisfying $0\to \Bbb{Z} \oplus \Bbb{Z}/3\Bbb{Z} \to G \to \Bbb{Z} \oplus \Bbb{Z}/3\Bbb{Z} \to 0$
QUESTION [23 upvotes]: I am studying for my qualifying exams and came across the following question: 


Find all abelian groups $G$ that fit into an exact sequence $0\to \Bbb{Z} \oplus \Bbb{Z}/3\Bbb{Z} \to G \to \Bbb{Z} \oplus \Bbb{Z}/3\Bbb{Z} \to 0$.


From another question on this site, I know I can make a diagram  of the form
$$\require{AMScd}
\begin{CD}
0 @>>> \mathbb{Z}^2 @>>> \mathbb{Z}^4 @>>> \mathbb{Z}^2 @>>> 0 \\
@. @VVV @VVV @VV V @. \\
0 @>>> \mathbb{Z} \oplus \Bbb{Z}/3\Bbb{Z} @>>> G @>>> \mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z} @>>> 0.
\end{CD}
$$
Let us call the three vertical maps, from left to right respectively $f$, $g$ and $h$. By the Snake Lemma, I can extend the diagram above to one that looks like
\begin{CD}
 @. 0 @>>>0  @>>>0 @. \\
@. @VVV @VVV @VV V @. \\
0 @>>> \ker f @>>> \ker g @>>> \ker h @>>> 0 \\
@. @VVV @VVV @VV V @. \\
0 @>>> \mathbb{Z}^2 @>>> \mathbb{Z}^4 @>>> \mathbb{Z}^2 @>>> 0 \\
@. @VVV @VVV @VV V @. \\
0 @>>> \mathbb{Z} \oplus \Bbb{Z}/3\Bbb{Z} @>>> G @>>> \mathbb{Z}\oplus \mathbb{Z}3\mathbb{Z} @>>> 0\\
@. @VVV @VVV @VV V @. \\
 @. 0 @>>>0  @>>>0 @. \\
\end{CD}


Question 1: Since $\ker f = \ker h = \Bbb{Z}$, and $\Bbb{Z}$ is projective I know that the top non-zero row splits, and so $\ker g \cong \Bbb{Z}^2$. Can I say that $\ker f \hookrightarrow \ker g$ is inclusion into the first factor, and $\ker g\to \ker h$ is projection onto the second? 


If this is true, then I can present $G$ as the cokernel of the relations matrix
$$\left(\begin{array}{cc}  0 & a \\ 3 & b \\ 0 & 0 \\ 0 & 3  \end{array}\right)$$
for some integers $a,b\in\Bbb{Z}.$ This comes from analyzing the far top right and top left squares, and of course the assumption that question 1 is true.


Question 2: How can I simplify this matrix to give me nice information on what $G$ should be?


Edit 1: From the statement of the splitting lemma on pg. 147 of Hatcher, it looks as if I can basically arrange so that what I want in question 1 is true.

REPLY [6 votes]: It is clear that $G$ is finitely generated and as noted above it has free rank two. It suffices to determine what $\tau(G)$ (the torsion part of $G$) can be. Now $\tau(G)={\rm Tor}(G,\Bbb Q/\Bbb Z)$ which is left exact so we get an exact sequence $$0\longrightarrow \Bbb Z_3\longrightarrow \tau(G)\stackrel{f}\longrightarrow \Bbb Z_3\stackrel{g}\longrightarrow \Bbb Q/\Bbb Z\longrightarrow  (\Bbb Q/\Bbb Z)^2\longrightarrow \Bbb Q/\Bbb Z\longrightarrow 0 $$ 
since the torsion part of any group dies when tensoring with the divisible group $\Bbb Q/\Bbb Z$, and the free part contributes as many copies as its rank.
Since $\Bbb Z_3$ is simple, truncating this (i.e. inserting ${\rm im}(f)\longrightarrow 0$ at the end) means $\tau(G)$ fits into an exact sequence of the form $$0\longrightarrow \Bbb Z_3 \longrightarrow \tau (G) \longrightarrow  0$$ or $$ 0 \longrightarrow  \Bbb Z_3 \longrightarrow \tau(G) \longrightarrow \Bbb Z_3 \longrightarrow 0$$
which gives $\tau(G)$ either $\Bbb Z_3,\Bbb Z_9, \Bbb Z_3^2$.