1

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

KON3
  • 4,111
  • Is it specified whether $a$ and $b$ share any factors? – preferred_anon Dec 28 '21 at 15:25
  • Surely the second and third isomorphisms are nonsense and irrelevant anyway? – ancient mathematician Dec 28 '21 at 15:31
  • 2
    Rather than get confused with all these $a$s and $b$s it seems to me easier to prove once and for all that if $M$ is a normal subgroup of $X$ and $N$ is a normal subgroup of $Y$ then $M\times N$ is a normal subgroup of $X\times Y$ and moreover $X\times Y/M\times N$ is isomorphic to $X/M\times Y/N$. – ancient mathematician Dec 28 '21 at 15:34
  • It's better to write quotients in the form $G/H$ rather than $\frac{G}{H}$. This is because quotients are defined in terms of cosets. – Shaun Dec 28 '21 at 16:02
  • @ancientmathematician Brilliant answer,. I was searching for this hint only. Thank you, my problem is solved. :-) – KON3 Dec 28 '21 at 17:09

1 Answers1

3

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,

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

Shaun
  • 44,997