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I'm asked to classify all quotient groups of $\mathbb Z^2$ of order $24$.

So, a quotient of an abelian group is abelian, and since $\mathbb Z^2$ is finitely generated as an abelian group, so does any of its quotients. By the classification theorem, any finitely generated abelian group is one of $C_3\times C_8, C_3\times C_2\times C_4$, or $C_3\times C_2\times C_2\times C_2$. The first group is the quotient of $\mathbb Z^2$ by $3\mathbb Z\times 8\mathbb Z$. The second group is the quotient by $6\mathbb Z\times 4\mathbb Z$. Is that correct? What about the third group?

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  • $\begingroup$ Hint: $C_2\times C_2$ is the Klein four group $\endgroup$
    – janmarqz
    Jun 14, 2018 at 0:56
  • $\begingroup$ @janmarqz I'm not sure how I can use this. $\endgroup$
    – user557
    Jun 15, 2018 at 22:17

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Yes what you have is correct, it follows from the fact that a quotient of a cartesian product by a product of respective subgroups is the product of the individual quotients.

The third group doesn't have a presentation with two generators (consequence of the structure theorem for finite type abelian groups and the three factors of order $2 $). Therefore it cannot be a quotient of a group generated by two elements.

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  • $\begingroup$ Thanks. Could you clarify how to deduce that the third group doesn't have a presentation with two generators? The theorem says that any finite finitely generated abelian group is a product of cyclic groups of prime power orders. Also I'm not sure how to justify the statement in your last sentence. $\endgroup$
    – user557
    Jun 14, 2018 at 1:59
  • $\begingroup$ Yes, and these factors can be merged and split using the Chinese Remainder Theorem, and that's it. Since $2 $ is not coprime with itself, you can't merge these factors together. Now a presentation with two generators would necessarily imply that $ G $ is a product of two cyclic groups (presentations of finite type abelian groups can be simplified in a way that preserves the number of generators). As for the last sentence, a quotient also preserves the number of generators, it only ads relations. $\endgroup$ Jun 14, 2018 at 2:20
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    $\begingroup$ Would it be correct to say this? Identify $C_2$ with $\mathbb Z_2$. The generators of the group $C_2\times C_2\times C_2$ are the generators of the $\mathbb Z$-module $C_2\times C_2\times C_2$. This module can be also regarded as $\mathbb Z_2$-module (because the only non-zero coefficients in linear combinations of basis vectors are 0 and 1) and hence as a $\mathbb Z_2$-vector space. If it were generated by 2 elements, it would have dimension 2 over $\mathbb Z_2$, but $C_2\times C_2\times C_2$ has dimension 3. So $C_2\times C_2\times C_2$ cannot be generated by 2 elements. $\endgroup$
    – user557
    Jun 15, 2018 at 21:44
  • $\begingroup$ @user437307 It is correct and very nice! $\endgroup$ Jun 16, 2018 at 1:30

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