Inverse (finite group) isomorphism of a certain form exists I have been working on something in group theory for a long time and I have one problem that I cannot solve. I have reduced that problem to a conjecture. It takes some work to set it up, but I don't thinks it's that hard to understand.
Let $\mathcal G$ be the finite abelian group
$\sum_\limits{i=1}^A \mathbb Z/a_i \mathbb Z.$
We represent an element 
$\bar{\mathbf x} \in \mathcal G$
as a column vector
$\begin{bmatrix}
  \overline{x_1} &
  \overline{x_2} &
  \overline{x_3} &
  \cdots &
  \overline{x_A}
\end{bmatrix}^\text T$
where
$\mathbf x =
\begin{bmatrix}
  x_1 &
  x_2 &
  x_3 &
  \cdots &
  x_A
\end{bmatrix}^\text T \in \mathbb Z^{A \times 1}$
Define
$D_a = \operatorname{diag}(a_1, a_2, a_3, \dots, a_A)$.
Then, for any two
$\overline{\mathbf x}, \overline{\mathbf y} \in \mathcal G, \,$
$\overline{\mathbf x} = \overline{\mathbf y}$ if and only if there exists a
$\mathbf u \in \mathbb Z^{A \times 1}$
such that
$\mathbf x = \mathbf y + D_a \mathbf u$.
Let $\mathcal H$ be a nontrivial subgroup of $\mathcal G$ which can be expressed as an internal direct sum of cycles, 
$$\mathcal H =
  \bigoplus_{j=1}^B \langle \, \overline{\mathbf s_j} \, \rangle.$$
For $j=1 .. B,$ let
$\operatorname{ord}(\, \overline{\mathbf s_j} \,) = b_j$.
Let $S =
\begin{bmatrix}
  \mathbf s_1 &
  \mathbf s_2 &
  \mathbf s_3 &
  \cdots &
  \mathbf s_B
\end{bmatrix} \in \mathbb Z^{A \times B}$
be the matrix formed by the column vectors
$\{\mathbf s_j\}_{j=1}^B$
Define
$D_b = \operatorname{diag}(b_1, b_2, b_3, \dots, b_B)$.
Then there exists a matrix
$\widetilde S \in \mathbb Z^{A \times B}$ such that 
$$ S D_b = D_a \widetilde S $$
This is true because each $b_j$ is the smallest positive integer such that, for every $i$, there exists an integer $\tilde s_{ij}$ such that
$b s_{ij} = \tilde s_{ij} a_i \equiv 0 \pmod{a_i}.$
Finally, we get to
Conjecture:
The matrix
$\begin{bmatrix} \widetilde S \\ -D_b \end{bmatrix}$
has an integer left inverse.
This is equivalent to saying that the rows of $\widetilde S$ span $\prod_{j=1}^B \mathbb Z_{b_j}$.
It is easy to show that the elements in each column of
$\begin{bmatrix} \widetilde S \\ -D_b \end{bmatrix}$
are relatively prime. But I can't see how to take the next step and create an integer left inverse.
Context:
The mapping
$f : \prod_{j=1}^B \mathbb Z_{b_j} \to \mathcal H$
defined by
$f(\bar{\mathbf x}) = \overline{S \mathbf x}$
is an (additive group) isomorphism.
So there must exists an isomorphism
$\, f^{-1} : \mathcal H \to \prod_{j=1}^B \mathbb Z_{b_j}$
It would be nice if there were a matrix
$T \in \mathbb Z^{B \times A}$
such that 
$f^{-1}(\bar{\mathbf y}) = \overline{T \mathbf y}$,
but this only happens some times. If my conjecture were true, then we would always have 
$T = D_b \widetilde T D_a^{-1} \in \mathbb Q^{B \times A}$
where, for some
$U \in \mathbb Z^{B \times B} $, 
$$\begin{bmatrix} \widetilde T & U \end{bmatrix}
\begin{bmatrix} \widetilde S \\ -D_b \end{bmatrix} = I$$
I tried for a long time to find a counter example with no success, but I also can't prove that
$\begin{bmatrix}
    \widetilde S \\
    -D_b \\
 \end{bmatrix}$
always has a left inverse.
NOTES:
The expression 
$ S D_b = D_a \widetilde S $
is just the matrix way of expressing that
for $j = 1..B, \operatorname{ord}\mathbf s_j = b_j$
What the function
$F:\mathbb Z^B \to \mathcal H$
defined by 
$F(\bar{\mathbf x}) 
  = \overline{S x}
  = \sum_{j=1}^B x_j \mathbf{\bar s}_j$
is doing is pretty obvious. The problem is that the function is "too big" in the sense that any $x_j$ can be replaced by
$x_j + nb_j \; (n \in \mathbb Z)$
without altering the image of 
$\bar{\mathbf x}.$
That periodic redundancy is encapsulated in the function
$f : \prod_{j=1}^B \mathbb Z_{b_j} \to \mathcal H$
defined by
$f(\bar{\mathbf x}) = \overline{S \mathbf x}$.
Here is how you prove that this function is well-defined.
\begin{align}
  \bar{\mathbf x} = \bar{\mathbf y}
  &\implies x = y + D_b u \; (u \in \mathbb Z^{B \times 1})\\
  &\implies Sx = Sy + S D_b u\\
  &\implies Sx = Sy + D_a (\widetilde S u)\\
  &\implies f(\bar{\mathbf x}) = f(\bar{\mathbf y})\\
\end{align}
So $f$ is a well-defined isomorphism. Hence there is also an inverse isomorphism 
$f^{-1}:\mathcal H \to \prod_{j=1}^B \mathbb Z_{b_j}$.
Let's assume that there is a rational matrix
$T \in \mathbb Q^{B \times A}$
such that
$f^{-1}(\bar{\mathbf y}) =\overline{Ty}.$
We will show that $f^{-1}$ is well-defined if there exists an integer matrix
$\widetilde T \in \mathbb Z^{B \times A}$ such that
$T D_a = D_b \widetilde T$.
\begin{align}
  \bar{\mathbf y} = \bar{\mathbf z}
  &\implies y = z + D_a u \; (u \in \mathbb Z^{A \times 1})\\
  &\implies Ty = Tz + T D_a u\\
  &\implies Ty = Tz + D_b (\widetilde T u)\\
  &\implies f^{-1}(\bar{\mathbf y}) = f^{-1}(\bar{\mathbf z})\\
\end{align}
We also need to ensure that, for $j = 1..B$,
$f^{-1}(f(\bar{\mathbf e}_j)) = f^{-1}(\bar{\mathbf s}_j) 
 = \bar{\mathbf e}_j.$
This can all be compressed into
$f^{-1}(f(\overline{I_B})) = \overline{I_B}$.
\begin{align}
  f^{-1}(f(\overline{I_B})) &= \overline{I_B} \\
  f^{-1}(\overline S) &= \overline{I_B} \\
  \overline{TS} &= \overline{I_B} \\
  TS &= I_B + D_b U \; (U \in \mathbb Z^{B \times B})\\
  (D_b \widetilde T D_a^{-1})(D_a \widetilde S D_b^{-1}) &= I_B + D_b U \\
  D_b \widetilde T \widetilde S D_b^{-1} &= I_B + D_b U \\
  \widetilde T \widetilde S &= I_B + U D_b \\
  \widetilde T \widetilde S - U D_b &= I_B \\
  \begin{bmatrix} \widetilde T & U \end{bmatrix}
  \begin{bmatrix} \widetilde S \\ -D_b \end{bmatrix} &= I_B
\end{align}
 A: This is at best a partial answer, but, I am posting it here because it is more answer than it is comment and because it takes up a lot of space to present. I have observed that the key is to get down to the level of p-groups.
Consider the group 
$\mathcal G = \mathbb Z_{35} \times \mathbb Z_{42} \times\mathbb Z_{80}$,with elements
$\mathbf s_1 = 
\begin{bmatrix}  
  1 \\ 21 \\ 0
\end{bmatrix}$
and
$\mathbf s_2 = 
\begin{bmatrix}  
  0 \\ 30 \\ 1
\end{bmatrix}$
of orders $70$ and $560$ respectively. We break $\mathcal G$ down as follows.
$$
\begin{array}{ccccc}
\mathbb Z_{35} & \times & \mathbb Z_{42} & \times & \mathbb Z_{80} \\
[x &  & y &  & z]^{\text T}
\end{array}
$$
$$ \longleftrightarrow $$
$$
\begin{array}{ccccccccccccc}
(\mathbb Z_{2} & \times & 
\mathbb Z_{16}) & \times & 
(\mathbb Z_{3}) & \times &
(\mathbb Z_{5} & \times & 
\mathbb Z_{5}) & \times &
(\mathbb Z_{7} & \times &
\mathbb Z_{7})\\
[y & & z & & y & & x & & z & & x & & y]^{\text T}
\end{array}
$$
Then
$$ 
\mathbf S = 
\begin{bmatrix}
   1 &  0\\
  21 & 30\\
   0 &  1
\end{bmatrix}
\leftrightarrow
\begin{bmatrix}
  21 &  0 \\
   0 &  1 \\
  21 &  0 \\
   1 &  0 \\
   0 &  1 \\
   1 &  0 \\
  21 & 30
\end{bmatrix}
\leftrightarrow
\begin{bmatrix}
   1 &  0 \\
   0 &  1 \\
   0 &  0 \\
   1 &  0 \\
   0 &  1 \\
   1 &  0 \\
   0 &  2
\end{bmatrix}
$$
$$
\left[\begin{array}{c|c|c|c}
A &Ab &\cdots &A^{n-1}b
\end{array}\right]
$$
So $\widetilde S = D_a^{-1} S D_b
=\begin{bmatrix}
\dfrac 12 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & \dfrac{1}{16} & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & \dfrac 13 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & \dfrac 15 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & \dfrac 15 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & \dfrac 17 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & \dfrac 17 \\
\end{bmatrix}
\begin{bmatrix}
   1 &  0 \\
   0 &  1 \\
   0 &  0 \\
   1 &  0 \\
   0 &  1 \\
   1 &  0 \\
   0 &  2
\end{bmatrix}
\begin{bmatrix}
  70 & 0 \\
   0 & 560
\end{bmatrix}
=
\begin{bmatrix}
   35 &   0 \\
    0 &  35 \\
    0 &   0 \\
   14 &   0 \\
    0 & 112 \\
   10 &   0 \\
    0 & 160
\end{bmatrix}$
