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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.

  • 0
    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](http://en.wikipedia.org/wiki/Direct_sum_of_modules#Universal_property) $\mathbf Z_4 \oplus \mathbf Z_{16} = \mathbf Z_4 \times \mathbf Z_{16}$.2012-03-13
  • 0
    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).2012-03-13

1 Answers 1