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Here's an exercise on my homework assignment:

Let $G = (\mathbb{Z} \times \mathbb{Z}) \times \mathbb{Z}$ under componentwise addition. Let $D$ be the cyclic subgroup of $\mathbb{Z} > \times \mathbb{Z}$ generated by $(-2,3)$. Determine the isomorphism class of the finitely generated abelian group $G/(D\times > 4\mathbb{Z})$. Carefully prove your answer using the first isomorphism theorem.

The theorem:

Let $\phi:G\to G'$ be a group homomorphism with kernel $H$. Then $\phi > [G]$ is a group, and $\mu:G/H\to\phi[G]$ given by $\mu(gH)=\phi(g)$ is an isomorphism. If $\gamma:G\to G/H$ is the homomorphism given by $\gamma(g)=gH$, then $\phi(g)=\mu \gamma (g)$ for each $g\in G$.

I think $G/(D\times 4\mathbb{Z})$ is isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_4$. I determined this by setting $H=(D\times 4\mathbb{Z})$ and considering the factor group $G/H$, which I determined to be $a_iH$ for $1\le i \le 12$ where $a_1,a_2,...,a_{12} = (0,0,0), (0,0,1), (0,0,2), (0,0,3), (0,1,0,), (0,1,1), (0,1,2), (0,1,3), (0,2,0), (0,2,1), (0,2,2), (0,2,3)$ respectively. Taking these $a_i$'s as the representative elements of $G/H$, it is clear that $G/H$ is isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_4$.

First of all, is this a valid way to show that $G/(D \times 4\mathbb{Z})$ is isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_4$?

Second of all, how do I prove this with the first isomorphism theorem?

Here's the example given in the book of using the first isomorphism theorem:

Problem: Show that $(\mathbb{Z}_4 \times \mathbb{Z}_2)/(\{0\}\times > \mathbb{Z}_2)$ is isomorphic to $\mathbb{Z}_4$.

Solution: The projection map $\pi:\mathbb{Z}_4 \times \mathbb{Z}_2 \to \mathbb{Z}_4$ given by $\pi(x,y)=x$ is a homomorphism of $\mathbb{Z}_4 \times > \mathbb{Z}_2$ onto $\mathbb{Z}_4$ with kernel $\{0\}\times > \mathbb{Z}_2$. By the first isomorphism theorem, we know that the given factor group is isomorphic to $\mathbb{Z}_4$

Comparing this example to the theorem, it looks like $\pi=\phi$, $\mathbb{Z}_4 \times \mathbb{Z}_2=G$, $\mathbb{Z}_4 = G'$, and $\{0\} \times \mathbb{Z}_2$ = $H$. It looks like all I need to do to prove that $G/(D \times 4\mathbb{Z})$ is isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_4$ is find the homomorphism $\phi: G\to G'$ s.t. the kernel of $\phi$ is $D \times 4\mathbb{Z}$, but I have no idea how to come up with this homomorphism.

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  • $\begingroup$ $((\mathbb{Z} \times \mathbb{Z}) \times \mathbb{Z})/(D \times 4\mathbb{Z}) \cong ((\mathbb{Z} \times \mathbb{Z})/D) \times \mathbb{Z}/4\mathbb{Z} \cong ((\mathbb{Z} \times \mathbb{Z})/D) \times \mathbb{Z}_4$. So you have to recognize the group $(\mathbb{Z} \times \mathbb{Z})/D.$ Why this group is isomorphic to $\mathbb{Z}_3?$ It's actually an infinite group (finitely generated rank one $\mathbb{Z}$-module.) $\endgroup$ – Krish Nov 21 '14 at 2:47
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A general fact (can be proved using first isomorphism theorem):

Let $A_1, A_2$ be two groups and let $B_i \subseteq A_i$ be a normal subgroup for $i = 1, 2.$ Then $B_1 \times B_2$ is a normal subgroup of $A_1 \times A_2$ and $(A_1 \times A_2)/(B_1 \times B_2) \cong (A_1/B_1) \times (A_2/B_2).$

Using the above fact, we get: $$ ((\mathbb{Z} \times \mathbb{Z}) \times \mathbb{Z})/(D \times 4\mathbb{Z}) \cong ((\mathbb{Z} \times \mathbb{Z})/D) \times (\mathbb{Z} /4\mathbb{Z}) \cong ((\mathbb{Z} \times \mathbb{Z})/D) \times \mathbb{Z}_4. $$ So we have to study the group $(\mathbb{Z} \times \mathbb{Z})/D.$

Define a map $\phi : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$ by $(a, b) \mapsto 3a + 2b.$ Then $\phi$ is a group homomorphism. It is surjective and ker$\phi = D.$ So by first isomorphism theorem, $(\mathbb{Z} \times \mathbb{Z})/D \cong \mathbb{Z}.$

Combining all these we get: $((\mathbb{Z} \times \mathbb{Z}) \times \mathbb{Z})/(D \times 4\mathbb{Z}) \cong \mathbb{Z} \times \mathbb{Z}_4.$

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