# How to show that $Z/12Z×Z/90Z×Z/25Z$ and $Z/100Z×Z/30Z×Z/9Z$ are isomorph?

How to show that $G = \Bbb Z/12\Bbb Z \times \Bbb Z/90 \Bbb Z\times \Bbb Z/25 \Bbb Z$ and $H = \Bbb Z/100 \Bbb Z \times \Bbb Z/30\Bbb Z\times \Bbb Z/9\Bbb Z$ are isomorph?

The way I would go is to use the decomposition in primary groups:

For G:

• $12 = 2^2 * 3$
• $90 = 3^2*2*5$
• $25 = 5^2$

So $G \simeq G(2) \times G(3) \times G(5)$ with

• $G(2)= (\Bbb Z/2\Bbb Z) \times (\Bbb Z/2^2\Bbb Z)$
• $G(3)= (\Bbb Z/3\Bbb Z) \times (\Bbb Z/3^2\Bbb Z)$
• $G(5)= (\Bbb Z/5\Bbb Z) \times (\Bbb Z/5^2\Bbb Z)$

For H:

• $100 = 5^2 * 2^2$
• $30 = 5*3*2$
• $9 = 3^2$

So $H \simeq H(2) \times H(3) \times H(5)$ with

• $H(2)= (\Bbb Z/2\Bbb Z) \times (\Bbb Z/2^2\Bbb Z)$
• $H(3)= (\Bbb Z/3\Bbb Z) \times (\Bbb Z/3^2\Bbb Z)$
• $H(5)= (\Bbb Z/5\Bbb Z) \times (\Bbb Z/5^2\Bbb Z)$

As $G$ and $H$ have the same decomposition they are isomorph.

Is it right? Is there a better/faster/easier way to prove it?

• That’s pretty much the way I’d have done it. Jun 17 '18 at 14:03

Your proof using the primary decomposition is fine and systematic.

You can also use the Chinese remainder theorem to recombine the factors into the invariant factor decomposition:

$G = \Bbb Z/12\Bbb Z \times \Bbb Z/90 \Bbb Z\times \Bbb Z/25 \Bbb Z \cong \Bbb Z/3\Bbb Z \times \Bbb Z/4\Bbb Z \times \Bbb Z/9 \Bbb Z \times \Bbb Z/10 \Bbb Z\times \Bbb Z/25 \Bbb Z \cong \Bbb Z/30\Bbb Z\times \Bbb Z/900 \Bbb Z$

$H = \Bbb Z/100 \Bbb Z \times \Bbb Z/30\Bbb Z\times \Bbb Z/9\Bbb Z \cong \Bbb Z/30\Bbb Z\times \Bbb Z/100 \Bbb Z \times \Bbb Z/9\Bbb Z \cong \Bbb Z/30\Bbb Z\times \Bbb Z/900 \Bbb Z$

• Interesting, thanks. I have trouble understanding the invariant factor decomposition, Wikipedia is not very explicit, any resource to understand it better? Like how to choose which $Z/nZ$ you recombine together or not ? Jun 17 '18 at 14:20
• @astudentofmaths, $H$ was easy and then I made $G$ follow $H$.
– lhf
Jun 17 '18 at 21:47

I would have displayed it this way.

\begin{align} \mathbb Z_{12} × \mathbb Z_{90} × \mathbb Z_{25} &\cong (\mathbb Z_4 \times \mathbb Z_3) \times (\mathbb Z_2 \times \mathbb Z_9 \times \mathbb Z_5) \times (\mathbb Z_{25}) \\ &\cong (\mathbb Z_4 \times \mathbb Z_{25}) \times (\mathbb Z_2 \times \mathbb Z_3 \times \mathbb Z_5) \times (\mathbb Z_9) \\ &\cong \mathbb Z_{100} × \mathbb Z_{30} × \mathbb Z_9 \end{align}