Quotient Group Classification

Question

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.

Answer 1

$\DeclareMathOperator{\bZ}{\mathbf Z}$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 $\bZ/2\bZ \times \bZ/8\bZ$.