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I needed help in classifying the following quotient groups according to the fundamental theorem of finitely generated abelian groups:

$$ \begin{array} &(\mathbb Z_4 \times \mathbb Z_{16})/\langle(1, 4)\rangle,\\ (\mathbb Z_4 \times\mathbb Z_{16})/\langle(2, 4)\rangle,\\ (\mathbb Z \times \mathbb Z \times \mathbb Z)/\langle(1, 2, 4)\rangle. \end{array} $$

What I tried out:

(i) $F : \mathbb Z_4 \times \mathbb Z_{16}\longrightarrow \mathbb Z_{16}$ defined by $F(a, b) = 4a - b \mod 16$ is a well-defined surjective homomorphism with $\ker F = \langle(1, 4)\rangle.$

$F$ is well-defined: Writing $a + 4j,$ and $b + 16k$ for any integers $j, k,$ we have $$4(a + 4j) - (b + 16k) = (4a - b) + 16(j-k) = 4a - b \mod 16$$ for any $j, k.$

$F$ is a homomorphism: For any $(a, b), (c, d)$ in $\mathbb Z_4 \times \mathbb Z_{16},$ we have

$$F(a, b) + F(c, d) = (4a - b) + (4c - d) = 4(a + c) - (b + d) = F(a+c, b+d).$$

$F$ is surjective: For any $c$ in $\mathbb Z_{16},$ we have $$F(0, -c) = 4 \cdot 0 - (-c) = c,$$ as required.

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  • $\begingroup$ What you have looks good. You should have a remark about why the kernel is what you say it is. Another way of organizing this is to note that $\mathbf Z_{16}$ has a unique subgroup of order $4$, generated by $4 \bmod{16}$, so there is an injective homomorphism $\mathbf Z_4 \to \mathbf Z_{16}$ sending $(1 \bmod4)$ to $(4 \bmod{16})$. And of course we have negation $\mathbf Z_{16} \to \mathbf Z_{16}$. Now use the universal property of the direct sum $\mathbf Z_4 \oplus \mathbf Z_{16} = \mathbf Z_4 \times \mathbf Z_{16}$. $\endgroup$ – Dylan Moreland Mar 13 '12 at 0:04
  • $\begingroup$ What you haven't shown is that the kernel is precisely the subgroup generated by $(1,4)$; it is reasonably clear that $(1,4)$ is contained in the kernel; you have to explain why the subgroup it generates equals the kernel. This can be done explicitly (show that if $(a,b)\in\mathrm{ker}(F)$ then $(a,b) = (r,4r)$ for some $r$), or via a counting argument (you know that $\mathbb{Z}_4\times\mathbb{Z}_{16}/\mathrm{ker}(F) \cong \mathbb{Z}_{16}$, so the kernel must have order $4$; since it contains $\langle(1,4)\rangle$, which is of order $4$, the latter equals the kernel). $\endgroup$ – Arturo Magidin Mar 13 '12 at 1:22
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$\DeclareMathOperator{\bZ}{\mathbf Z}$To add to my comments above: I'm not clever enough to always come up with a good map. However, you can use Smith normal form, which comes up often on this site, to do the other two parts mindlessly [But you should try to understand the math behind it; I won't try to explain that here.]. For example, in the second part we have a surjective homomorphism \[ \bZ \times \bZ \to \bZ/4\bZ \times \bZ/16\bZ. \] The kernel $H$ is generated by $(4, 0)$ and $(0, 16)$. And the preimage of $\langle(2, 4)\rangle$ is generated by $(2, 4)$ and $H$. So put the three generators as the columns of a relations matrix \[ \begin{pmatrix} 4 & 0 & 2 \\ 0 & 16 & 4 \end{pmatrix} \] which, assuming that I've done this correctly, we can place into the following form using row and column operations: \[ \begin{pmatrix} 2 & 0 & 0 \\ 0 & 8 & 0 \end{pmatrix}, \] from which it follows that your quotient is $\bZ/2\bZ \times \bZ/8\bZ$.

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