# Quotients of order $24$ of a free abelian group

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?

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

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.
• 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. Jun 14, 2018 at 2:20
• 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. Jun 15, 2018 at 21:44