Number of subgroup $G<\Bbb Z^3$ such that $\Bbb Z^3/G\simeq \Bbb Z/3\Bbb Z\oplus\Bbb Z/3\Bbb Z$ [closed]

Compute the number of subgroups $G<\Bbb Z^3$ such that $\Bbb Z^3/G\simeq \Bbb Z/3\Bbb Z\oplus\Bbb Z/3\Bbb Z$.

Possible forms of $G$ are $G = \Bbb Z\oplus 3\Bbb Z\oplus 3\Bbb Z$, $3\Bbb Z\oplus 3\Bbb Z\oplus\Bbb Z$ and $3\Bbb Z\oplus\Bbb Z\oplus 3\Bbb Z$. Is there anything more?

Let’s start with a simpler case. Suppose you were asked to count subgroups $G$ of $\mathbb{Z}\oplus\mathbb{Z}$ such that $(\mathbb{Z}\oplus\mathbb{Z})/G\cong \mathbb{Z}/2\mathbb{Z}$.

Your answer suggests two options: $2\mathbb{Z}\oplus\mathbb{Z}$ and $\mathbb{Z}\oplus\mathbb{2Z}$. But is that all? Instead of looking at quotients of subgroups, let’s think of the implied homomorphism $f: \mathbb{Z}\oplus\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$.

Such a homomorphism is determined by the images of a generating set, for which we can take $(1,0)$ and $(0,1)$. Between them, those images must generate $\mathbb{Z}/2\mathbb{Z}$, but otherwise we are free to choose them. Once we pick them, we get a kernel $G=\ker(f)$ and the required isomorphism will hold.

Your proposed solution identifies the homomorphisms $f((1,0))=1, f((0,1))=0$ and $f((1,0))=0,f((0,1))=1$. Those two are great, but you have another option: $f((1,0))=f((0,1))=1$. The kernel of this homomorphism is a subgroup $G$ made up of those vectors $(a,b)$ such that $a\equiv b\pmod{2}$. That’s another legitimate subgroup which is neither of the groups you suggested.

You can apply this approach to the question on hand. Pick a convenient generating set, determine where its elements must be mapped by a surjective homomorphism, and count the kernels. You’ll see more than $3$ possibilities.