1. ## Direct Product Isomorphism

Could anyone please explain clearly, why is it that we have $\mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{3^2} \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5$, but $\mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \ncong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5$?

So, why is it that $\mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5$ is isomorphic to the first direct product but not to the second one?

I mean, we can break down $\mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5$ to the following (which is the same as the first direct product):

$\mathbb{Z}_3 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3^2}\oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

And we further to the following (which is the same as the second direct product):

$\mathbb{Z}_3 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

So is it not isomorphic to the latter?

2. Originally Posted by demode
Could anyone please explain clearly, why is it that we have $\mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{3^2} \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5$, but $\mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \ncong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5$?

So, why is it that $\mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5$ is isomorphic to the first direct product but not to the second one?

I mean, we can break down $\mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5$ to the following (which is the same as the first direct product):

$\mathbb{Z}_3 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3^2}\oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

And we further to the following (which is the same as the second direct product):

$\mathbb{Z}_3 \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

So is it not isomorphic to the latter?
In effect, you are asking why $\mathbb{Z}_{3^2}$ is not isomorphic to $\mathbb{Z}_{3} \oplus \mathbb{Z}_{3}$. One answer would be that the first of those groups has elements of order 9, but all the elements (except the identity) in the second group have order 3.

In general, it is true that $\mathbb{Z}_{mn}$ is isomorphic to $\mathbb{Z}_{m} \oplus \mathbb{Z}_{n}$ if m and n are co-prime, but not otherwise.

3. I see, but here's another example I don't understand:

" $\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}$. But $\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \ncong \mathbb{Z}_{60}$."

So for the first one we know that $\mathbb{Z}_{30} \cong \mathbb{Z}_3 \oplus \mathbb{Z}_{10}$, but can we write

$\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10}$?

My notes say that if $m=n_1n_2...n_k$, then $Z_m$ is isomorphic to $Z_{n_1} \oplus ... \oplus Z_{n_k}$ iff $n_i$ and $n_j$ are coprime when $i \neq j$.

But in this case $2$ and $10$ are not coprime. Furthermore in $\mathbb{Z}_6 \oplus \mathbb{Z}_{10}$, 6 and 10 are not coprime either.

So how did they conclude $\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}$?

4. Originally Posted by demode
I see, but here's another example I don't understand:

" $\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}$. But $\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \ncong \mathbb{Z}_{60}$."

So for the first one we know that $\mathbb{Z}_{30} \cong \mathbb{Z}_3 \oplus \mathbb{Z}_{10}$, but can we write

$\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10}$?

My notes say that if $m=n_1n_2...n_k$, then $Z_m$ is isomorphic to $Z_{n_1} \oplus ... \oplus Z_{n_k}$ iff $n_i$ and $n_j$ are coprime when $i \neq j$.

But in this case $2$ and $10$ are not coprime. Furthermore in $\mathbb{Z}_6 \oplus \mathbb{Z}_{10}$, 6 and 10 are not coprime either.

So how did they conclude $\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}$?
It requires you to note two things, that gcd(10, 3)=1 and gcd(2, 3)=1, so you get,

$\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}$.

5. Originally Posted by Swlabr
It requires you to note two things, that gcd(10, 3)=1 and gcd(2, 3)=1, so you get,

$\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}$.
Why is it that you don't require that gcd(2,10)=1? Those two are not co-prime! I thought they ALL should be co-prime with each other, at least that's what my notes say...

6. You only need that the one's you're actually manipulating be coprime: Notice that $\mathbb{Z}_2$ never interacts directly with $\mathbb{Z}_{10}$

7. Originally Posted by demode
Originally Posted by Swlabr
It requires you to note two things, that gcd(10, 3)=1 and gcd(2, 3)=1, so you get,

$\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{10} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}$.
Why is it that you don't require that gcd(2,10)=1? Those two are not co-prime! I thought they ALL should be co-prime with each other, at least that's what my notes say...
It might make it more obvious if you write it like this:

$\mathbb{Z}_2 \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \bigl(\mathbb{Z}_3 \oplus \mathbb{Z}_{10}\bigr) \cong \bigl(\mathbb{Z}_2 \oplus \mathbb{Z}_3\bigr) \oplus \mathbb{Z}_{10} \cong \mathbb{Z}_6 \oplus \mathbb{Z}_{10}$.

That makes it clear that only co-prime numbers are being used at each step. It also brings out a hidden point in the proof, namely that the direct sum operation is associative.