<s>[INST] Let $G=\mathbb{Z}/24\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$. Consider the quotient group $H=G/\langle (10,3,2)\rangle$. Determine a direct product of cyclic groups that is isomorphic to $H$.
The Smith Normal form can be used to find the invariant factors in the structure theorem for finitely generated abelian groups. I set up the following matrix $$\begin{pmatrix} 24 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 3 \\ 10 & 3 & 2\end{pmatrix}$$ but I was unable to find the SMF of this matrix.
I know the following about the structure of $H$:
$|\langle (10,3,2)\rangle|=12$
Also $|G|=24*6*3$ so by Lagrange's Theorem, $|H| = 36$.
By the fundamental theorem of finitely generated abelain groups, $H$ may have the following structure:
$\mathbb{Z}/36\mathbb{Z}$
$\mathbb{Z}/18\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$
$\mathbb{Z}/12\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$
$\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$
I am not sure how to identify the correct group structure without the invariant factors from the SMNF.
Edit: Here is some of my work by hand. Am I proceeding correctly?
\begin{align*}
  &\,
  \begin{pmatrix}
    24 & 0 & 0 \\
     0 & 6 & 0 \\
     0 & 0 & 3 \\
    10 & 3 & 2
  \end{pmatrix}
  \sim
  \begin{pmatrix}
    10 & 3 & 2 \\
    24 & 0 & 0 \\
     0 & 6 & 0 \\
     0 & 0 & 3
  \end{pmatrix}
  \sim
  \begin{pmatrix}
    2 & 10 & 3 \\
    0 & 24 & 0 \\
    0 &  0 & 6 \\
    3 &  0 & 0
  \end{pmatrix}
  \\
  \sim&\,
  \begin{pmatrix}
    2 & 10 &  1 \\
    0 & 24 &  0 \\
    0 &  0 &  0 \\
    3 &  0 & -3
  \end{pmatrix}
  \sim
  \begin{pmatrix}
     1 & 2 & 10 \\
     0 & 0 & 24 \\
     0 & 0 &  0 \\
    -3 & 3 &  0
  \end{pmatrix}
  \sim
  \begin{pmatrix}
    1 & 2 & 10 \\
    0 & 0 & 24 \\
    0 & 0 &  0 \\
    0 & 9 & 30
  \end{pmatrix}
  \\
  \sim&\,
  \begin{pmatrix}
    1 & 0 &  0 \\
    0 & 0 & 24 \\
    0 & 0 &  0 \\
    0 & 9 & 30
  \end{pmatrix}
  \sim
  \begin{pmatrix}
    1 & 0 &  0 \\
    0 & 9 & 30 \\
    0 & 0 & 24 \\
    0 & 0 &  0
  \end{pmatrix}
\end{align*}
(Original image here.)
 [/INST] The Smith Form has diagonal entries 1, 3, 12 so that the quotient is isomorphic to $Z_3\times Z_{12}$.
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