ra.rings and algebras - equivalence of submodules

It is enough to show that

if $Msubseteq mathbb Z^3$ be a subgroup such that $mathbb Z^3/M$ is a cyclic group of order $k$, then there exists $ginmathrm{SL}(3,mathbb Z)$ such that $g(M)=langle (1,0,0),(0,1,0),(0,0,k)rangle$.

Let $Msubseteq mathbb Z^3$ be a subgroup such that $mathbb Z^3/M$ is a cyclic group of order $k$. Then $M$ is free of rank $3$, and there exists $Ain M(3,mathbb Z)$ such that $M=Acdotmathbb Z^3$. Using the Smith normal form, we know that there exists $3times 3$ matrices $P$ and $Q$, invertible over $mathbb Z$, such that $PAQ=D$ with $D=left(begin{smallmatrix}a\&b\&&cend{smallmatrix}right)$ and $amid bmid c$.
Then $PM=PAQmathbb Z^3=Dmathbb Z^3$.

It follows that $Pinmathrm{SL}(3,mathbb Z)$ is such that $PM$ is generated by $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$ with $amid bmid c$. Since $mathbb Z^3/g(M)$ is cyclic of order $k$, we must have $a=b=1$ and $c=k$. This tells us that the claim above is true.

(I've done everything at the level of generality which your problem needs, and I'll leave the fun of finding the correct general statement for you...)