Justify $\frac{\mathbb{Z}_a\times \mathbb{Z}_b}{\langle a/c\rangle \times \langle b/d\rangle}$ is isomorphic to $\mathbb{Z}_c\times \mathbb{Z}_d$.

Question

Let $c\mid a, d\mid b$ in $\mathbb{N}$. Then the external direct product $\mathbb{Z}_a\times \mathbb{Z}_b$ has a subgroup $\langle a/c\rangle \times \langle b/d\rangle$ of order $cd$.

To examine whether or not $\frac{\mathbb{Z}_a\times \mathbb{Z}_b}{\langle a/c\rangle \times \langle b/d\rangle}$ is isomorphic to $\mathbb{Z}_c\times \mathbb{Z}_d$.

I once posted a similar question here however this is not the present case. Below is the work I tried.

From the book Contemporary Abstract Algebra written by Gallian, I am aware that if $k$ is a divisor of $n$ then $\mathbb{Z}_n/\langle k\rangle \simeq \mathbb{Z}_k$. However I tried to show the above claim like this: $$\frac{\mathbb{Z}_a\times \mathbb{Z}_b}{\langle a/c\rangle \times \langle b/d\rangle}\simeq \frac{\mathbb{Z}_a}{\langle a/c\rangle}\times \frac{\mathbb{Z}_b}{\langle b/d\rangle}\simeq \langle a/c\rangle \times \langle b/d\rangle \simeq \mathbb{Z}_c\times \mathbb{Z}_d.$$

Is it correct? If not, how can I modify the mistake ? Or whether the claim above is not true in general? Thanks in well advance.

Edit As a further conclusion, $$\frac{\mathbb{Z}_a\times \mathbb{Z}_b}{\mathbb{Z}_c\times \mathbb{Z}_d}\simeq \mathbb{Z}_{a/c}\times \mathbb{Z}_{b/d}$$

Answer 1

Theorem: Let $M\unlhd G$ and $N\unlhd H$ as groups. Then $$(G\times H)/(M\times N)\cong (G/M)\times (H/N).$$

Proof: Let

$$\begin{align} \varphi: G\times H&\to (G/M)\times (H/N)\\ (g,h)&\mapsto (gM, hN). \end{align}$$

Clearly $\varphi$ is a well-defined, surjective homomorphism. (Why?)

Then $\ker(\varphi)=M\times N$.

Now, by the First Isomorphism Theorem,

$$(G\times H)/(M\times N)\cong (G/M)\times (H/N).\,\square$$