# equivalence of submodules

I have Z^3/M = Z^3/N = Z_k where M,N are submodules of Z^3 and Z_k is cyclic order k.

I would like to say some SL_3(Z) transformation takes M to N. Is this true? How to show?

-

It is enough to show that

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

Let $M\subseteq \mathbb Z^3$ 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 $A\in M(3,\mathbb Z)$ such that $M=A\cdot\mathbb Z^3$. Using the Smith normal form, we know that there exists $3\times 3$ matrices $P$ and $Q$, invertible over $\mathbb Z$, such that $PAQ=D$ with $D=\left(\begin{smallmatrix}a\\\&b\\\&&c\end{smallmatrix}\right)$ and $a\mid b\mid c$. Then $PM=PAQ\mathbb Z^3=D\mathbb Z^3$.

It follows that $P\in\mathrm{SL}(3,\mathbb Z)$ is such that $PM$ is generated by $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$ with $a\mid b\mid c$. 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...)

-
Just a comment, Mariano -- it's much awesomer, when we get questions that are "too localized" like this, if you give an extremely general answer, than if you, ahem, "do someone's homework". –  Scott Morrison Feb 12 '10 at 5:13
This answer was very helpful, particularly pointing me to Smith normal form. Should I be explaining how this isn't homework? Or disguising my question in a form that's more general? –  AndrewLMarshall Feb 12 '10 at 12:11
@mathuni, oh, probably just ignore me, I was being grumpy, sorry. I guess the better advice is: when you're asking a question that could be mistaken as homework, provide a little more context, background or motivation, just so we can all recognize the question immediately as legitimate use of the site. Many people here feel that "homework questions" are abuse of the site, and react poorly. I shouldn't have in this case, but did. –  Scott Morrison Feb 12 '10 at 16:39
I understand completely. This little speed bump came up as I'm trying to classify a certain set of symplectic toric orbifolds by looking at primitive vectors at vertices of the moment polytope in Z^3. Of course, none of that's relevant to solving this bit here, but I see why stating the context makes it more legitimate. –  AndrewLMarshall Feb 12 '10 at 23:35