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.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.