Taking quotients over isometric groups and determining the cokernel of transformations

Question

We know that given group $A$ and subgroups $B_1,B_2$ such that $B_1\cong B_2$, it is not necessary that $A/B_1\cong A/B_2$. Now take for instance the following transformation $\varphi:\mathbb{Z}^3\rightarrow \mathbb{Z}^3$ $$ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix} $$ Because column and row operations preserve isometry, we can reduce it to the normal smith form $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ Similarly to vector spaces, this allows us to determine the kernel and the image: $\mathrm{Im} (\varphi)\cong \mathbb{Z}\oplus3\mathbb{Z}\cong \mathbb{Z}^2$. Intuitively I know that the cokernel is supposed to be $\mathbb{Z}^3/\mathbb{Z}\oplus3\mathbb{Z}=\mathbb{Z}/3\oplus\mathbb{Z}$ and not $\mathbb{Z}^3/\mathbb{Z}^2=\mathbb{Z}$, but I don't understand why is it true mathematically. I understand that the normal smith form is supposed to represent in some way what the group structure 'actually' is. But haven't we only stated that it preserves the group's isometry?