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Determine, and identify, the subgroups of $G=\mathbb Z^2$ that contain the smaller subgroup $H=2\mathbb Z \times 4\mathbb Z$.

Note: I would like to identify subgroups not just determine them like in another example one determines a subgroup of $\mathbb Z^2$ that contains $2\mathbb Z \times 2\mathbb Z$ to be $\{(2k+n,2l+n)|k,l,n \in \mathbb Z\} = \{(a,b)|a \equiv b \bmod 2\}$.

Here's what I tried. Question 1: What is wrong or missing here?

  1. I think $G/H=\{(a,b)+H|a=0,1;b=0,1,2,3\}$

    • Question 2: In general, is $\frac{\mathbb Z^2}{m\mathbb Z \times n\mathbb Z} = \{(a,b)+H|a=0,1, ..., m-1;b=0,1, ..., n-1\}$?

  2. $G/H$ is isomorphic to $\mathbb Z_2 \times \mathbb Z_4$ by $\gamma: \mathbb Z_2 \times \mathbb Z_4 \to G/H, \gamma(a,b)=(a,b)+H$

  3. Ummm... The following is the set of subgroups of $\mathbb Z_2 \times \mathbb Z_4$: $\{\langle(a,b)\rangle|a=0,1;b=0,1,2,3\}$...

    • Question 3: Okay so now as I type it up I realise where I'm going wrong. What do I have to do: really check each of the 8x8=64 pairs $(a,b)$, $(c,d)$ to see if I get another subgroup? And then continue with 8x8x8=512 triples $(a,b)$, $(c,d)$, $(e,f)$ to see if I get another subgroup? Well by correspondence theorem...am I looking for only 8 subgroups total?

(But anyway I'll just type up what I got before the above realisation.)

  1. By (2) and (3), the subgroups of $G/H$ are $\{\langle (a,b) + H \rangle|a=0,1;b=0,1,2,3\}$.

  2. By correspondence theorem, the subgroups of $G$ that contain $H$ are $\{\pi_H^{-1} (\langle (a,b) + H \rangle) = H + \langle (a,b) \rangle|a=0,1;b=0,1,2,3\}$

    • 5.1. Re that, I'm assuming that there's no $\bmod$ for '$H + \langle (a,b) \rangle$' with $H + \langle (a,b) \rangle$ as really a sum/product set, but there is $\bmod$ for '$\langle (a,b) + H \rangle$' with $(a,b) + H = (a,b) \mod H$.

Finally, we have, for $\langle (a,b) \rangle = \{p(a,b)|p \in \mathbb Z\}$ and $H=\{(2m,4n)|m,n \in \mathbb Z\}$, that:

  1. $H + \langle (0,1) \rangle = \{(2m,4n+p)|m,n,p \in \mathbb Z\} = 2 \mathbb Z \times \mathbb Z?$
  2. $H + \langle (0,2) \rangle = \{(2m,2(2n+p))|m,n,p \in \mathbb Z\} = 2 \mathbb Z \times 2 \mathbb Z?$
  3. $H + \langle (0,3) \rangle = \{(2m,4n+3p)|m,n,p \in \mathbb Z\} = 2 \mathbb Z \times \mathbb Z$ again?
  4. $H + \langle (1,0) \rangle = \{(2m+p,4n)|m,n,p \in \mathbb Z\} = \mathbb Z \times 4 \mathbb Z?$
  5. $H + \langle (1,1) \rangle = \{(2m+p,4n+p)|m,n,p \in \mathbb Z\}$: Question 4: How do I identify this? I tried $\{(a,b)|b\equiv a \bmod 2\}$, but this is what I got for the $(1,3)$ below, yet I think the 2 subgroups are different.
  6. $H + \langle (1,2) \rangle = \{(2m+p,2(2n+p))|m,n,p \in \mathbb Z\} = \{(a,b)|b \equiv 2a \bmod 4\} ?$
  7. $H + \langle (1,3) \rangle = \{(2m+p,4n+3p)|m,n,p \in \mathbb Z\}$: Question 5: How do I identify this? I tried $\{(a,b)|b\equiv 3a \bmod 2\}$, but I think $\{(a,b)|b\equiv 3a \bmod 2\} = \{(a,b)|b\equiv a \bmod 2\}$ and then this is what I got for the $(1,1)$ above, yet I think the 2 subgroups are different.
  8. $H + \langle (0,0) \rangle = H$
  9. Missing: $\mathbb Z \times 2\mathbb Z$?
BCLC
  • 13,459
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    It might help a little to use the fact that every subgroup of $\mathbb{Z}^2$ can be written as the subgroup generated by at most two elements. Therefore, the same also holds for subgroups of $\mathbb{Z}^2 / (2\mathbb{Z} \times 4\mathbb{Z})$. – Daniel Schepler Oct 21 '21 at 17:36
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    Well, if you know the theory of Smith normal form it isn't too hard: any subgroup of $\mathbb{Z}^2$ is finitely generated so it's the image of some $2 \times n$ matrix $A$. Then you can find $P \in GL_2(\mathbb{Z})$, $Q \in GL_n(\mathbb{Z})$, and diagonal $D \in M_{2\times n}(\mathbb{Z})$ such that $A = PDQ$. Then the image of $A$ is generated by $d_{11} \cdot P e_1$ and $d_{22} \cdot P e_2$. – Daniel Schepler Oct 21 '21 at 18:47

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