Finding missing entries to commutative $3 \times 3$ diagram with exact rows and columns 
Assume that the following diagram of abelian groups has exact rows and columns. Can you determine the missing entries and maps? Give short reasoning.
  $$
  \require{AMScd}
  \begin{CD}
    {}
    @.
    0
    @.
    0
    @.
    0
    {}
    {}
    \\
    @.
    @VVV
    @VVV
    @VVV
    \\
    0
    @>>>
    {}
    @>>>
    {}
    @>>>
    \mathbb{Z}/2\mathbb{Z}
    @>>>
    0
    \\
    @.
    @VVV
    @VVV
    @VVV
    \\
    0
    @>>>
    {}
    @>>>
    \mathbb{Z}
    @>>>
    {}
    @>>>
    0
    \\
    @.
    @VVV
    @VVV
    @VVV
    \\
    0
    @>>>
    \mathbb{Z}/3\mathbb{Z}
    @>>>
    {}
    @>>>
    \mathbb{Z}/2\mathbb{Z}
    @>>>
    0
    \\
    @.
    @VVV
    @VVV
    @VVV
    \\
    {}
    @.
    0
    @.
    0
    @.
    0
    {}
    {}
  \end{CD}
$$
  (Original image of this diagram here.)

This is what I tried:
$$
  \require{AMScd}
  \begin{CD}
    {}
    @.
    0
    @.
    0
    @.
    0
    {}
    {}
    \\
    @.
    @VVV
    @VVV
    @VVV
    \\
    0
    @>>>
    3\mathbb{Z}
    @> a >>
    6\mathbb{Z}
    @> b >>
    \mathbb{Z}/2\mathbb{Z}
    @>>>
    0
    \\
    @.
    @V g VV
    @V h VV
    @V i VV
    \\
    0
    @>>>
    \mathbb{Z}
    @> c >>
    \mathbb{Z}
    @> d >>
    \mathbb{Z}/2\mathbb{Z}
    @>>>
    0
    \\
    @.
    @V j VV
    @V k VV
    @V l VV
    \\
    0
    @>>>
    \mathbb{Z}/3\mathbb{Z}
    @> e >>
    \mathbb{Z}/6\mathbb{Z}
    @> f >>
    \mathbb{Z}/2\mathbb{Z}
    @>>>
    0
    \\
    @.
    @VVV
    @VVV
    @VVV
    \\
    {}
    @.
    0
    @.
    0
    @.
    0
    {}
    {}
  \end{CD}
$$
(Original image of this diagram here.)
Where for example I let $a$ be given by $t \mapsto 4t$ and $b$ be given by $t \mapsto t/6$. I defined all the other functions in similar ways such that all rows and all columns became exact, however I missed the crucial part that the diagram must be commutative and everything failed. I need some hints on this one please.
 A: Okay I don't know how to make commutative diagrams on this site, but you know that the second row, third column must have $4$ elements since it is in between two sets of two elements, so it is either $(\mathbb{Z}/2\mathbb{Z})^2$ or $\mathbb{Z}/4\mathbb{Z}$. Since $\mathbb{Z}$ must surject onto it from the left, it can only be $\mathbb{Z}/4\mathbb{Z}$. 
From this we know that the second row, first column must be $4\mathbb{Z}$, since it is the kernel of the map $\mathbb{Z} \rightarrow \mathbb{Z}/4\mathbb{Z}: x \rightarrow x \mod(4)$. The natural map $4\mathbb{Z} \rightarrow \mathbb{Z}/3\mathbb{Z}$ takes an element $x \rightarrow x \mod(12)$, so in the first column, first row we should have $12\mathbb{Z}$. 
Since we need an exact sequence $0 \rightarrow 12\mathbb{Z} \rightarrow ? \rightarrow \mathbb{Z}/2\mathbb{Z}$, by the same logic we have been using we get $6\mathbb{Z}$ in the first row, second column. Finally we need the kernel of the inclusion map $6 \mathbb{Z} \rightarrow \mathbb{Z}$ in the third row, second column, which is $\mathbb{Z}/6\mathbb{Z}$.
